Equation of State of a Perfect Gas: Ideal Gas Laws, Universal Gas Constant, Dalton’s Law & JEE Problems | Class 11 Physics

Welcome to Lecture 101 of our comprehensive JEE Physics Master Series! Today, we launch Unit 9: Kinetic Theory of Gases. This unit forms the microscopic bridge connecting classical Newtonian mechanics with macroscopic thermodynamics. We begin with the foundational macroscopic description: the Equation of State of a Perfect (Ideal) Gas and a rigorous recap of the classical empirical gas laws (Boyle’s, Charles’s, Gay-Lussac’s, and Avogadro’s laws). We analyze the universal gas constant $R$ and Boltzmann constant $k_B$, alternative equations of state in terms of mass density and number density, Dalton’s Law of Partial Pressures for non-reacting gas mixtures, graphical transformations and slope comparisons across indicator planes, and high-yield JEE Advanced problem archetypes including connecting flasks, barometric lapse rates, and faulty barometer capillary columns.

Detailed Concept Card: Equation of State & Classical Gas Laws

1. The Ideal Gas Model & Microscopic Assumptions

An ideal (perfect) gas is a theoretical physical model characterized by two fundamental microscopic postulates:

1. Negligible Molecular Volume: Gas molecules are treated as point masses whose finite physical volume is negligibly small compared to the total volume $V$ occupied by the gas ($V_{\text{molecules}} \ll V_{\text{container}}$).
2. Zero Intermolecular Forces: The molecules exert no mutual attractive or repulsive forces on one another except during instantaneous, perfectly elastic collisions. Consequently, the internal energy of an ideal gas is purely translational and rotational/vibrational kinetic energy; potential energy of interaction is zero ($U = K_{\text{internal}}$).

When do Real Gases Behave as Ideal Gases?
Real gases (such as $\text{He}, \text{H}_2, \text{N}_2, \text{O}_2$) closely approximate ideal gas behavior under conditions of low pressure ($P \to 0$) and high temperature ($T \gg T_c$).
– At low pressure: Molecules are widely separated, rendering their finite volume negligible compared to the container volume.
– At high temperature: Molecular kinetic energies are enormous compared to weak intermolecular van der Waals attractions, rendering potential energy negligible.

2. Classical Empirical Gas Laws Recap

The equation of state synthesizes four foundational empirical gas laws discovered experimentally between the 17th and 19th centuries:

Gas Law	Constant Variables	Mathematical Relation	Key Graphical & Physical Feature
Boyle’s Law (Robert Boyle, 1662)	Temperature ($T = \text{const}$), Quantity ($n = \text{const}$)	$$P \propto \frac{1}{V} \implies P V = \text{constant}$$ $$P_1 V_1 = P_2 V_2$$	Isotherms on $P$-$V$ are rectangular hyperbolas. On a $P$ vs. $1/V$ plot, isotherms are straight lines through the origin whose slope is $nRT$. On $PV$ vs. $P$, isotherms are horizontal straight lines.
Charles’s Law (Jacques Charles, 1787)	Pressure ($P = \text{const}$), Quantity ($n = \text{const}$)	$$V \propto T \implies \frac{V}{T} = \text{constant}$$ $$\frac{V_1}{T_1} = \frac{V_2}{T_2} \quad (T \text{ in K})$$	Centigrade form: $V_t = V_0(1 + \alpha_v t)$ where $\alpha_v = \frac{1}{273.15}\text{ }^\circ\text{C}^{-1}$. Volume extrapolates to zero at Absolute Zero ($-273.15^\circ\text{C} = 0\text{ K}$). Isobar slope on $V$-$T$ is $\frac{nR}{P}$ (steeper line $\iff$ lower pressure).
Gay-Lussac’s / Amontons’s Law (1802)	Volume ($V = \text{const}$), Quantity ($n = \text{const}$)	$$P \propto T \implies \frac{P}{T} = \text{constant}$$ $$\frac{P_1}{T_1} = \frac{P_2}{T_2} \quad (T \text{ in K})$$	Centigrade form: $P_t = P_0(1 + \alpha_p t)$ where $\alpha_p = \frac{1}{273.15}\text{ }^\circ\text{C}^{-1}$. Pressure coefficient equals volume coefficient ($\alpha_p = \alpha_v$). Isochore slope on $P$-$T$ is $\frac{nR}{V}$ (steeper line $\iff$ smaller volume).
Avogadro’s Law (Amedeo Avogadro, 1811)	Pressure ($P = \text{const}$), Temperature ($T = \text{const}$)	$$V \propto n \implies \frac{V}{n} = \text{constant}$$	Equal volumes of all gases under identical conditions of temperature and pressure contain an equal number of molecules. Standard Molar Volume at STP ($273.15\text{ K}, 1\text{ atm}$) is $V_m \approx 22.414\text{ L/mol}$ ($22.7\text{ L/mol}$ at $1\text{ bar}$).

3. The Unified Equation of State of a Perfect Gas

Combining Boyle’s, Charles’s, and Avogadro’s laws yields the master Ideal Gas Equation of State:

$$\mathbf{P V = n R T = \left(\frac{m}{M}\right) R T = N k_B T}$$

where the variables and fundamental constants are:

• $P$: Absolute pressure of the gas ($\text{N/m}^2$ or $\text{Pa}$, where $1\text{ atm} = 1.01325 \times 10^5\text{ Pa} = 760\text{ mmHg} = 76\text{ cmHg} = 1.013\text{ bar}$).
• $V$: Volume occupied by the gas ($\text{m}^3$, where $1\text{ L} = 10^{-3}\text{ m}^3 = 1000\text{ cm}^3$).
• $T$: Absolute thermodynamic temperature in Kelvin ($T = t(^\circ\text{C}) + 273.15$).
• $n$: Number of moles of gas ($n = \frac{m}{M} = \frac{N}{N_A}$).
• $m$: Total mass of the gas sample; $M$: Molar mass ($\text{kg/mol}$ or $\text{g/mol}$).
• $N$: Total number of gas molecules; $N_A \approx 6.02214 \times 10^{23}\text{ mol}^{-1}$ is Avogadro’s constant.
• $R$: Universal Gas Constant, identical for all gases: $$\mathbf{R = 8.314\text{ J/(mol}\cdot\text{K)} = 1.987 \approx 2.0\text{ cal/(mol}\cdot\text{K)} = 0.08206\text{ L}\cdot\text{atm/(mol}\cdot\text{K)} = 8.314 \times 10^7\text{ erg/(mol}\cdot\text{K)}}$$
• $k_B$: Boltzmann Constant (gas constant per single molecule): $$\mathbf{k_B = \frac{R}{N_A} = \frac{8.314}{6.022 \times 10^{23}} \approx 1.3806 \times 10^{-23}\text{ J/K}}$$

Alternative Representations of the Equation of State

1. In terms of Mass Density ($\rho = m/V$): $$P V = \left(\frac{m}{M}\right) R T \implies P = \frac{\rho R T}{M} \iff \mathbf{\rho = \frac{P M}{R T}}$$ Key Takeaway: At constant temperature, gas density is directly proportional to pressure ($\rho \propto P$). For a given pressure and temperature, heavier gases are denser ($\rho \propto M$).
2. In terms of Specific Gas Constant ($r = R/M$): $$P = \rho r T \quad \text{where } r \text{ depends on the specific chemical identity of the gas.}$$
3. In terms of Number Density ($n_0 = N/V$): $$\mathbf{P = n_0 k_B T} \quad \text{where } n_0 \text{ is the number of molecules per unit volume.}$$

4. Dalton’s Law of Partial Pressures for Gas Mixtures

For a mixture of $k$ non-reacting ideal gases enclosed in a common container of volume $V$ maintained at temperature $T$, each gas expands to fill the entire container volume and exerts a pressure independent of the other gases.

Dalton’s Law:
“The total pressure exerted by a mixture of non-reacting ideal gases is equal to the sum of the partial pressures that each constituent gas would exert if it alone occupied the entire volume of the container at the same temperature.”

$$\mathbf{P_{\text{total}} = \sum_{i=1}^k P_i = \sum_{i=1}^k \frac{n_i R T}{V} = \frac{R T}{V} \sum_{i=1}^k n_i = n_{\text{total}} \frac{R T}{V}}$$

The partial pressure ($P_i$) of the $i$-th gas is directly proportional to its mole fraction ($x_i$):

$$P_i = x_i P_{\text{total}} \quad \text{where } x_i = \frac{n_i}{n_{\text{total}}} = \frac{n_i}{\sum n_j} \quad \left(\sum x_i = 1\right)$$

The equivalent (apparent) molar mass ($M_{\text{mix}}$) of the gas mixture is:

$$\mathbf{M_{\text{mix}} = \frac{m_{\text{total}}}{n_{\text{total}}} = \frac{\sum m_i}{\sum n_i} = \frac{\sum n_i M_i}{\sum n_i} = \sum_{i=1}^k x_i M_i}$$

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Four Graded Solved Examples

Example 1 (JEE Main Level: Connecting Flasks at Different Temperatures)

Problem Statement: Two rigid glass vessels $A$ and $B$ of volumes $V_A = 3.0\text{ L}$ and $V_B = 2.0\text{ L}$ are connected by a thin tube of negligible volume fitted with a stopcock. Initially, the stopcock is closed: Vessel $A$ contains an ideal gas at pressure $P_A = 5.0\text{ atm}$ and temperature $T_A = 300\text{ K}$, while Vessel $B$ contains the same gas at pressure $P_B = 2.0\text{ atm}$ and temperature $T_B = 400\text{ K}$. The stopcock is opened, allowing gas to flow between the vessels until pressure equalizes. Throughout the process, Vessel $A$ is maintained at $300\text{ K}$ and Vessel $B$ is maintained at $400\text{ K}$. Determine the final common equilibrium pressure $P_f$ in the system (in $\text{atm}$) and the final number of moles in each vessel.

Solution:

Step 1: Calculate Initial Moles in Each Vessel.
Using the ideal gas law $n = \frac{PV}{RT}$:
In Vessel $A$: $$n_A = \frac{P_A V_A}{R T_A} = \frac{(5.0\text{ atm})(3.0\text{ L})}{R (300\text{ K})} = \frac{15}{300 R} = \frac{0.050}{R}$$ In Vessel $B$: $$n_B = \frac{P_B V_B}{R T_B} = \frac{(2.0\text{ atm})(2.0\text{ L})}{R (400\text{ K})} = \frac{4.0}{400 R} = \frac{0.010}{R}$$ Total moles in the closed composite system (strictly conserved): $$n_{\text{total}} = n_A + n_B = \frac{0.050}{R} + \frac{0.010}{R} = \frac{0.060}{R}$$

Step 2: Express Final Equilibrium State.
When pressure equalizes at $P_f$, the vessels remain at their respective constant temperatures ($T_A = 300\text{ K}, T_B = 400\text{ K}$): $$n_A’ = \frac{P_f V_A}{R T_A} = \frac{P_f \times 3.0}{300 R} = 0.010 \frac{P_f}{R}$$ $$n_B’ = \frac{P_f V_B}{R T_B} = \frac{P_f \times 2.0}{400 R} = 0.005 \frac{P_f}{R}$$ Total final moles: $$n_{\text{total}} = n_A’ + n_B’ = (0.010 + 0.005) \frac{P_f}{R} = 0.015 \frac{P_f}{R}$$

Step 3: Solve for Equilibrium Pressure $P_f$.
Equating initial and final total moles: $$0.015 \frac{P_f}{R} = \frac{0.060}{R} \implies P_f = \frac{0.060}{0.015} = \mathbf{4.0\text{ atm}}$$ Final mole distribution: $$n_A’ = 0.010 \times \frac{4.0}{R} = \frac{0.040}{R} = \frac{2}{3} n_{\text{total}}$$ $$n_B’ = 0.005 \times \frac{4.0}{R} = \frac{0.020}{R} = \frac{1}{3} n_{\text{total}}$$

Takeaway: Because Vessel $A$ is colder ($300\text{ K}$ vs $400\text{ K}$), gas density is higher in $A$, so it holds twice as many moles as Vessel $B$ despite having only $1.5$ times the volume.

Example 2 (Graphical & Conceptual: Indicator Diagram Transformations & Slope Comparisons)

Problem Statement: A fixed mass of an ideal gas ($n\text{ moles}$) undergoes a thermodynamic cycle $1 \to 2 \to 3 \to 1$ represented on a $V$-$T$ indicator diagram:
– Path $1 \to 2$: Straight line passing through the origin from $(V_1, T_1)$ to $(2V_1, 2T_1)$.
– Path $2 \to 3$: Isothermal expansion at $2T_1$ from volume $2V_1$ to $4V_1$.
– Path $3 \to 1$: Straight line from $(4V_1, 2T_1)$ back to $(V_1, T_1)$.
1. Identify the nature of Path $1 \to 2$ and find the slope of its curve on the $V$-$T$ plane.
2. Determine the ratio of pressures $\frac{P_1}{P_2}$ and $\frac{P_2}{P_3}$.
3. Derive how the slope of an isobar on a mass density versus pressure ($\rho$-$P$) plot depends on temperature.

Solution:

Step 1: Nature and Slope of Path $1 \to 2$.
Because Path $1 \to 2$ is a straight line passing through the origin $(0, 0)$, $V \propto T$. By Charles’s Law, this represents an isobaric process ($P = \text{constant} = P_1$).
From $P V = n R T \implies V = \left(\frac{n R}{P_1}\right) T$, the slope on the $V$-$T$ plane is: $$\mathbf{\text{Slope}_{V-T} = \frac{n R}{P_1}}$$ Key deduction: The slope of an isobar on the $V$-$T$ plane is inversely proportional to pressure ($P \propto 1/\text{slope}$). Steeper lines correspond to lower pressures!

Step 2: Pressure Ratios.
– For Path $1 \to 2$ (isobaric): $P_2 = P_1 \implies \mathbf{\frac{P_1}{P_2} = 1}$.
– For Path $2 \to 3$ (isothermal at $T = 2T_1$): By Boyle’s Law ($P V = \text{const}$): $$P_2 (2V_1) = P_3 (4V_1) \implies P_3 = \frac{P_2}{2} \implies \mathbf{\frac{P_2}{P_3} = 2}$$ The pressure halves during the isothermal volume doubling.

Step 3: Density vs. Pressure ($\rho$-$P$) Slope Analysis.
From the equation of state: $$\rho = \left(\frac{M}{R T}\right) P$$ On a plot of density $\rho$ (vertical axis) versus pressure $P$ (horizontal axis), an isothermal process is a straight line through the origin with slope: $$\mathbf{\text{Slope}_{\rho-P} = \frac{d\rho}{dP} = \frac{M}{R T}}$$ As temperature increases, the slope decreases ($\text{Slope} \propto 1/T$). At higher temperatures, thermal agitation expands the gas, resulting in a lower mass density for the same applied pressure.

Example 3 (JEE Advanced Level: Atmospheric Scale Height & Barometric Formula)

Problem Statement: Assuming Earth’s atmosphere behaves as an ideal gas of uniform molar mass $M = 29.0\text{ g/mol}$ at a constant temperature $T_0 = 290\text{ K}$ under a uniform gravitational field $g = 9.80\text{ m/s}^2$ (an isothermal atmosphere):
1. Derive the barometric height formula expressing pressure $P(z)$ as a function of altitude $z$ above sea level, where $P(0) = P_0 = 1.013 \times 10^5\text{ Pa}$.
2. Calculate the atmospheric scale height $H = \frac{R T_0}{M g}$ (in $\text{km}$).
3. Find the altitude $z_{1/2}$ at which the atmospheric pressure drops to exactly half its sea-level value ($\ln 2 = 0.693$).

Solution:

Step 1: Hydrostatic Equilibrium and Derivation.
Consider a thin horizontal atmospheric air slice of thickness $dz$ and cross-sectional area $A$ at height $z$. Hydrostatic balance requires: $$P A – (P + dP) A = \rho g A dz \implies dP = -\rho g dz$$ From the ideal gas equation of state, density is $\rho = \frac{P M}{R T_0}$. Substituting: $$dP = -\left(\frac{P M}{R T_0}\right) g dz \implies \frac{dP}{P} = -\frac{M g}{R T_0} dz$$ Integrating from sea level ($z = 0, P = P_0$) to altitude $z$: $$\int_{P_0}^{P(z)} \frac{dP}{P} = -\frac{M g}{R T_0} \int_0^z dz \implies \ln\left(\frac{P(z)}{P_0}\right) = -\frac{M g z}{R T_0}$$ $$\mathbf{P(z) = P_0 \exp\left(-\frac{M g z}{R T_0}\right) = P_0 e^{-z / H}}$$ where $H = \frac{R T_0}{M g}$ is the scale height of the atmosphere.

Step 2: Calculate Scale Height $H$.
Convert molar mass to SI units: $M = 29.0\text{ g/mol} = 0.0290\text{ kg/mol}$. $$H = \frac{R T_0}{M g} = \frac{(8.314\text{ J/(mol}\cdot\text{K)})(290\text{ K})}{(0.0290\text{ kg/mol})(9.80\text{ m/s}^2)} = \frac{2411.06}{0.2842} \approx \mathbf{8484\text{ m} \approx 8.48\text{ km}}$$

Step 3: Altitude of Half Pressure ($z_{1/2}$).
Setting $P(z_{1/2}) = \frac{P_0}{2}$: $$\frac{P_0}{2} = P_0 e^{-z_{1/2} / H} \implies e^{z_{1/2} / H} = 2 \implies z_{1/2} = H \ln 2$$ $$z_{1/2} = 8484\text{ m} \times 0.693 \approx \mathbf{5879\text{ m} \approx 5.88\text{ km}}$$ At an elevation of approximately $5.88\text{ km}$ (higher than the base camp of Mount Everest), atmospheric pressure and density fall to $50\%$ of their sea-level values.

Example 4 (JEE Advanced Level: Inverted Capillary Tube with Trapped Air Column)

Problem Statement: A narrow uniform glass tube of length $L = 90.0\text{ cm}$ closed at one end contains a column of dry air trapped by an $h = 20.0\text{ cm}$ long pellet of mercury. When the tube is held vertically with the open end pointing upwards, the length of the trapped air column is $l_1 = 30.0\text{ cm}$. The external atmospheric pressure is $P_0 = 76.0\text{ cm of Hg}$. The tube is now slowly inverted in a vertical plane so that the open end points downwards. Assuming the temperature of the air column remains strictly constant, determine:
1. The new length $l_2$ of the trapped air column after inversion.
2. The length $l_3$ of the trapped air column when the tube is held horizontally.
3. Will any mercury spill out when the tube is held vertically open-end downwards?

Solution:

Step 1: Open End Upwards (Initial Vertical State 1).
The pressure of the trapped air column balances atmospheric pressure plus the weight of the mercury thread: $$P_1 = P_0 + h = 76.0\text{ cmHg} + 20.0\text{ cmHg} = \mathbf{96.0\text{ cmHg}}$$ Volume of the trapped air column of cross-sectional area $A$: $$V_1 = A \cdot l_1 = 30.0 \, A$$

Step 2: Open End Downwards (Inverted Vertical State 2).
Atmospheric pressure supports the trapped air pressure plus the mercury thread: $$P_2 + h = P_0 \implies P_2 = P_0 – h = 76.0\text{ cmHg} – 20.0\text{ cmHg} = \mathbf{56.0\text{ cmHg}}$$ Let $l_2$ be the new length of the air column ($V_2 = A \cdot l_2$). Since temperature is constant, by Boyle’s Law: $$P_1 V_1 = P_2 V_2 \implies (96.0\text{ cmHg})(30.0 \, A) = (56.0\text{ cmHg})(l_2 \, A)$$ $$l_2 = \frac{96.0 \times 30.0}{56.0} = \frac{2880}{56.0} = \frac{360}{7} \approx \mathbf{51.43\text{ cm}}$$

Step 3: Horizontal Orientation (State 3).
In horizontal orientation, the mercury column exerts no hydrostatic pressure along the tube axis: $$P_3 = P_0 = 76.0\text{ cmHg}$$ By Boyle’s Law ($P_3 V_3 = P_1 V_1$): $$l_3 = \frac{P_1 l_1}{P_3} = \frac{96.0 \times 30.0}{76.0} = \frac{2880}{76.0} = \frac{720}{19} \approx \mathbf{37.89\text{ cm}}$$

Step 4: Spill-Out Check.
When inverted open-end downwards, total length occupied by air plus mercury: $$l_{\text{total}} = l_2 + h = 51.43\text{ cm} + 20.0\text{ cm} = 71.43\text{ cm}$$ Since $l_{\text{total}} = 71.43\text{ cm} < L = 90.0\text{ cm}$, the entire mercury thread remains comfortably inside the tube. No mercury spills out.

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Practice Worksheet: 10 JEE Problems

Questions 1 to 5: JEE Main Level (Single Correct)

Q1. A real gas behaves most like an ideal gas under which of the following combinations of thermodynamic conditions?

(A) High pressure and high temperature
(B) Low pressure and low temperature
(C) Low pressure and high temperature
(D) High pressure and low temperature

Q2. According to Charles’s Law, when a sample of gas is cooled at constant pressure, its volume $V_t$ at temperature $t(^\circ\text{C})$ is related to its volume $V_0$ at $0^\circ\text{C}$ by $V_t = V_0(1 + \alpha_v t)$. The value of the volume expansion coefficient $\alpha_v$ is:

(A) $1/100\text{ }^\circ\text{C}^{-1}$
(B) $1/273.15\text{ }^\circ\text{C}^{-1}$
(C) $1/373.15\text{ }^\circ\text{C}^{-1}$
(D) $273.15\text{ }^\circ\text{C}^{-1}$

Q3. Two ideal gases $A$ and $B$ have molar masses $M_A$ and $M_B$ respectively, with $M_A = 2 M_B$. If the density of gas $A$ at temperature $T_A$ and pressure $P_A$ is equal to the density of gas $B$ at temperature $T_B$ and pressure $P_B = 2 P_A$, then the ratio of their absolute temperatures $\frac{T_A}{T_B}$ is:

(A) $1$
(B) $4$
(C) $1/4$
(D) $1/2$

Q4. Two rigid insulated vessels of volumes $V_1 = 2.0\text{ L}$ and $V_2 = 3.0\text{ L}$ containing the same ideal gas at pressures $P_1 = 4.0\text{ atm}$ and $P_2 = 1.0\text{ atm}$ respectively are connected by a valve. If the temperature of the entire system is maintained constant at $300\text{ K}$, the final equilibrium pressure after opening the valve is:

(A) $2.5\text{ atm}$
(B) $2.2\text{ atm}$
(C) $3.0\text{ atm}$
(D) $1.8\text{ atm}$

Q5. A closed rigid vessel contains a mixture of $4.0\text{ g}$ of Oxygen gas ($\text{O}_2$, molar mass $32\text{ g/mol}$) and $2.0\text{ g}$ of Hydrogen gas ($\text{H}_2$, molar mass $2.0\text{ g/mol}$) at temperature $T$. Assuming both gases behave ideally, the fraction of the total pressure exerted by Oxygen is:

(A) $1/9$
(B) $1/8$
(C) $2/3$
(D) $8/9$

Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)

Q6. For a fixed quantity of an ideal gas, which of the following statements regarding graphical representations is/are correct?

(A) On a $V$-$T$ indicator diagram, isobars are straight lines passing through the origin, and a line of greater slope corresponds to lower pressure.
(B) On a $P$-$T$ indicator diagram, isochores are straight lines passing through the origin, and a line of greater slope corresponds to smaller volume.
(C) On a plot of $\frac{PV}{T}$ versus pressure $P$, the graph is a horizontal straight line parallel to the pressure axis.
(D) On a plot of gas density $\rho$ versus pressure $P$, isotherms are straight lines passing through the origin, and a line of greater slope corresponds to higher temperature.

Q7. A gas mixture consists of $n_1$ moles of gas 1 (molar mass $M_1$) and $n_2$ moles of gas 2 (molar mass $M_2$) enclosed in volume $V$ at temperature $T$. Which of the following relations is/are correct?

(A) The partial pressure of gas 1 is $P_1 = \left(\frac{n_1}{n_1 + n_2}\right) P_{\text{total}}$.
(B) The apparent molar mass of the mixture is $M_{\text{mix}} = \frac{n_1 M_1 + n_2 M_2}{n_1 + n_2}$.
(C) The total density of the mixture is $\rho_{\text{mix}} = \frac{P_{\text{total}} M_{\text{mix}}}{R T}$.
(D) The total pressure is $P_{\text{total}} = (n_1 + n_2) \frac{RT}{V}$.

Q8. A uniform vertical glass tube closed at the bottom contains an air column trapped by a mercury thread of length $h$. The external atmospheric pressure is $P_0$. Which of the following operations will increase the length of the trapped air column?

(A) Tilting the tube from the vertical position at constant temperature.
(B) Inverting the tube so that the open end points downwards at constant temperature.
(C) Heating the trapped air while keeping the tube vertical.
(D) Accelerating the tube vertically upwards with constant acceleration $a$.

Questions 9 to 10: Numerical / Integer Answer Type

Q9. Two rigid containers $A$ and $B$ of volumes $V_A = 3.0\text{ L}$ and $V_B = 2.0\text{ L}$ are connected by a thin tube of negligible volume with a valve. Initially, container $A$ contains an ideal gas at $P_A = 5.0\text{ atm}$ and $T_A = 300\text{ K}$, while container $B$ contains the same gas at $P_B = 2.0\text{ atm}$ and $T_B = 400\text{ K}$. When the valve is opened, gas flows between the containers until a uniform pressure $P_f$ is established. If container $A$ is maintained at $300\text{ K}$ and container $B$ is maintained at $400\text{ K}$, calculate the final equilibrium pressure $P_f$ (in $\text{atm}$).

Q10. A faulty barometer tube of total length $L = 90\text{ cm}$ contains a small quantity of air trapped above the mercury column. When the true atmospheric pressure is $76\text{ cm of Hg}$, the barometer reads $70\text{ cm of Hg}$. At the same temperature, when the true atmospheric pressure falls to $71\text{ cm of Hg}$, find the reading of this faulty barometer (in $\text{cm of Hg}$).

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Detailed Solutions & Answer Key

Answer Key

Question	Correct Option / Answer	Topic / Concept
Q1	(C) Low pressure & high temperature	Real to Ideal Gas Transition Criteria
Q2	(B) $1/273.15\text{ }^\circ\text{C}^{-1}$	Charles’s Law Volume Expansion Coefficient
Q3	(A) $1$	Gas Density Ratio & Equation of State
Q4	(B) $2.2\text{ atm}$	Isothermal Connecting Vessels Mixture Pressure
Q5	(A) $1/9$	Dalton’s Law Partial Pressure Ratio
Q6	(A), (B), (C)	Indicator Diagram Slopes & Scaling
Q7	(A), (B), (C), (D)	Gas Mixtures & Apparent Molar Mass
Q8	(A), (B), (C)	Trapped Air Column Hydrostatics
Q9	4	Two-Temperature Connecting Flasks ($4\text{ atm}$)
Q10	66	Faulty Barometer Boyle’s Law Equilibrium ($66\text{ cmHg}$)

Complete Step-by-Step Solutions

Solution 1:
An ideal gas assumes negligible molecular volume and zero intermolecular attractive forces. Real gases satisfy these assumptions best at:
1. Low Pressure: Large intermolecular separation minimizes the fraction of container volume occupied by molecules.
2. High Temperature: High molecular kinetic energy overwhelms weak intermolecular potential attraction.
Therefore, a real gas approaches ideal behavior at low pressure and high temperature.
Correct Option: (C)

Solution 2:
Charles’s Law states that at constant pressure, the volume of a given mass of gas increases by $\frac{1}{273.15}$ of its volume at $0^\circ\text{C}$ for every $1^\circ\text{C}$ rise in temperature: $$V_t = V_0\left(1 + \frac{t}{273.15}\right) \implies \alpha_v = \frac{1}{273.15}\text{ }^\circ\text{C}^{-1}$$ Correct Option: (B)

Solution 3:
From the equation of state in terms of density: $$\rho = \frac{P M}{R T}$$ Given $\rho_A = \rho_B$: $$\frac{P_A M_A}{R T_A} = \frac{P_B M_B}{R T_B} \implies \frac{T_A}{T_B} = \left(\frac{P_A}{P_B}\right)\left(\frac{M_A}{M_B}\right)$$ Substitute $P_B = 2 P_A$ and $M_A = 2 M_B$: $$\frac{T_A}{T_B} = \left(\frac{1}{2}\right)(2) = \mathbf{1}$$ Correct Option: (A)

Solution 4:
Since temperature is uniform and constant at $T = 300\text{ K}$, Boyle’s law applies to the total moles: $$n_1 = \frac{P_1 V_1}{R T}, \quad n_2 = \frac{P_2 V_2}{R T}$$ $$n_{\text{total}} = n_1 + n_2 = \frac{P_1 V_1 + P_2 V_2}{R T}$$ After opening the valve, the total volume is $V_{\text{total}} = V_1 + V_2 = 2.0 + 3.0 = 5.0\text{ L}$: $$P_f = \frac{n_{\text{total}} R T}{V_{\text{total}}} = \frac{P_1 V_1 + P_2 V_2}{V_1 + V_2} = \frac{(4.0\text{ atm})(2.0\text{ L}) + (1.0\text{ atm})(3.0\text{ L})}{2.0 + 3.0} = \frac{8.0 + 3.0}{5.0} = \frac{11.0}{5.0} = \mathbf{2.2\text{ atm}}$$ Correct Option: (B)

Solution 5:
Calculate moles of each gas: $$n_{\text{O}_2} = \frac{4.0\text{ g}}{32\text{ g/mol}} = 0.125\text{ mol} = \frac{1}{8}\text{ mol}$$ $$n_{\text{H}_2} = \frac{2.0\text{ g}}{2.0\text{ g/mol}} = 1.000\text{ mol}$$ Total moles in the container: $$n_{\text{total}} = n_{\text{O}_2} + n_{\text{H}_2} = 0.125 + 1.000 = 1.125\text{ mol} = \frac{9}{8}\text{ mol}$$ By Dalton’s Law, the fraction of total pressure exerted by Oxygen is its mole fraction: $$\frac{P_{\text{O}_2}}{P_{\text{total}}} = x_{\text{O}_2} = \frac{n_{\text{O}_2}}{n_{\text{total}}} = \frac{1/8}{9/8} = \mathbf{\frac{1}{9}}$$ Correct Option: (A)

Solution 6:
(A) Correct: $V = \left(\frac{nR}{P}\right) T$. Slope is $\frac{nR}{P}$, which is inversely proportional to $P$. A steeper slope corresponds to lower pressure.
(B) Correct: $P = \left(\frac{nR}{V}\right) T$. Slope is $\frac{nR}{V}$, which is inversely proportional to $V$. A steeper slope corresponds to smaller volume.
(C) Correct: For an ideal gas, $\frac{PV}{T} = nR = \text{constant}$ at all pressures, yielding a horizontal straight line.
(D) Incorrect: $\rho = \left(\frac{M}{RT}\right) P$. The slope is $\frac{M}{RT}$, which is inversely proportional to temperature $T$. A steeper slope corresponds to lower temperature.
Correct Options: (A), (B), (C)

Solution 7:
(A) Correct: $P_1 = x_1 P_{\text{total}} = \left(\frac{n_1}{n_1 + n_2}\right) P_{\text{total}}$.
(B) Correct: $M_{\text{mix}} = \frac{m_{\text{total}}}{n_{\text{total}}} = \frac{n_1 M_1 + n_2 M_2}{n_1 + n_2}$.
(C) Correct: $\rho_{\text{mix}} = \frac{m_{\text{total}}}{V} = \frac{n_{\text{total}} M_{\text{mix}}}{V} = \frac{P_{\text{total}} M_{\text{mix}}}{R T}$.
(D) Correct: Ideal gas equation applied to the mixture gives $P_{\text{total}} = n_{\text{total}} \frac{RT}{V} = (n_1 + n_2) \frac{RT}{V}$.
Correct Options: (A), (B), (C), (D)

Solution 8:
Initially vertical with closed end at bottom: pressure inside is $P_1 = P_0 + \rho g h$.
(A) Correct: When tilted at angle $\theta$ to the vertical, the effective hydrostatic head becomes $h \cos\theta < h$. Pressure decreases to $P_0 + \rho g h \cos\theta$, so volume expands ($l$ increases).
(B) Correct: When inverted open end downwards, pressure decreases to $P_0 – \rho g h$, so volume expands ($l$ increases).
(C) Correct: By Charles’s Law at constant pressure, heating increases the volume of the trapped air column ($l \propto T$).
(D) Incorrect: Upward acceleration increases effective gravity to $g + a$, raising pressure to $P_0 + \rho(g + a)h$, which compresses the gas ($l$ decreases).
Correct Options: (A), (B), (C)

Solution 9:
Initial moles in $A$: $$n_A = \frac{P_A V_A}{R T_A} = \frac{5.0 \times 3.0}{300 R} = \frac{0.050}{R}$$ Initial moles in $B$: $$n_B = \frac{P_B V_B}{R T_B} = \frac{2.0 \times 2.0}{400 R} = \frac{0.010}{R}$$ Total moles $n_{\text{total}} = \frac{0.060}{R}$.
Final moles at common pressure $P_f$: $$n_A’ = \frac{P_f V_A}{R T_A} = \frac{3.0 P_f}{300 R} = 0.010 \frac{P_f}{R}$$ $$n_B’ = \frac{P_f V_B}{R T_B} = \frac{2.0 P_f}{400 R} = 0.005 \frac{P_f}{R}$$ Total final moles: $$n_A’ + n_B’ = 0.015 \frac{P_f}{R}$$ Equating initial and final moles: $$0.015 \frac{P_f}{R} = \frac{0.060}{R} \implies P_f = \frac{0.060}{0.015} = \mathbf{4\text{ atm}}$$ Answer: 4

Solution 10:
Total length of the barometer tube is $L = 90\text{ cm}$.
In Case 1: True pressure $P_1 = 76\text{ cmHg}$, faulty reading $h_1 = 70\text{ cmHg}$.
Pressure of trapped air: $$P_{\text{air}, 1} = P_1 – h_1 = 76 – 70 = 6\text{ cmHg}$$ Length of trapped air column: $$l_1 = L – h_1 = 90 – 70 = 20\text{ cm}$$ In Case 2: True pressure $P_2 = 71\text{ cmHg}$, let faulty reading be $h_2$.
Pressure of trapped air: $$P_{\text{air}, 2} = P_2 – h_2 = 71 – h_2$$ Length of trapped air column: $$l_2 = L – h_2 = 90 – h_2$$ Since temperature is constant, by Boyle’s Law: $$P_{\text{air}, 1} l_1 = P_{\text{air}, 2} l_2$$ $$6 \times 20 = (71 – h_2)(90 – h_2)$$ $$120 = 6390 – 161 h_2 + h_2^2$$ $$h_2^2 – 161 h_2 + 6270 = 0$$ Factoring the quadratic equation: $$(h_2 – 66)(h_2 – 95) = 0$$ Since the mercury column height must be less than the true atmospheric pressure ($h_2 < 71\text{ cmHg}$), the physical root is: $$h_2 = \mathbf{66\text{ cm of Hg}}$$ Answer: 66