Welcome to Lecture 107 of our comprehensive JEE Physics Master Series! Continuing in Unit 9: Kinetic Theory of Gases, we focus on the practical and analytical applications of statistical mechanics: Specific Heat Capacities of Gases ($C_p, C_v, \gamma$). Connecting molecular degrees of freedom to macroscopic thermodynamics, we analyze Mayer’s relation ($C_p – C_v = R$), trace the energetic mechanism distinguishing isochoric from isobaric heating, calculate heat capacities across molecular geometries, establish rigorous mixing formulas for multi-component gases, evaluate polytropic heat capacities ($C = C_v + \frac{R}{1 – n}$), and master high-yield JEE Advanced problem patterns.
Detailed Concept Card: Heat Capacities of Gases & Mayer’s Relation
1. Fundamental Definitions & The Physical Mechanism
Heat capacity quantifies the heat energy required to raise the temperature of a unit quantity of a substance by $1\text{ K}$ (or $1^\circ\text{C}$):
- Molar Heat Capacity at Constant Volume ($C_v$): The heat absorbed per mole of gas during an isochoric process ($\Delta V = 0$): $$C_v = \left(\frac{dQ}{n dT}\right)_v = \left(\frac{\partial U_m}{\partial T}\right)_v$$ Physical Insight: Because no boundary work is performed ($W = \int P dV = 0$), $100\%$ of the heat supplied directly increases the internal thermal energy ($\Delta Q_v = \Delta U$).
- Molar Heat Capacity at Constant Pressure ($C_p$): The heat absorbed per mole during an isobaric process ($\Delta P = 0$): $$C_p = \left(\frac{dQ}{n dT}\right)_p = \left(\frac{dU_m + P dV_m}{dT}\right)_p = C_v + P \left(\frac{\partial V_m}{\partial T}\right)_p$$ Physical Insight: At constant pressure, the gas must expand against the external atmosphere to maintain constant pressure. The heat supplied must both increase internal energy AND perform boundary expansion work ($\Delta Q_p = \Delta U + W$). Consequently, $C_p > C_v$ always.
2. Derivation of Mayer’s Relation
For one mole of an ideal gas, the equation of state is $P V_m = R T$. Differentiating with respect to $T$ at constant pressure:
$$P \left(\frac{\partial V_m}{\partial T}\right)_p = R$$
Substituting this directly into the First Law definition of $C_p$ gives the celebrated Mayer’s Relation:
$$\mathbf{C_p – C_v = R}$$
Specific Heat Formulation (Per Unit Mass):
If $c_p$ and $c_v$ denote specific heat capacities per unit mass (in $\text{J/(kg}\cdot\text{K)}$) and $M$ is the molar mass in $\text{kg/mol}$:
$$\mathbf{c_p – c_v = \frac{R}{M} = r}$$
where $r$ is the specific (individual) gas constant. Because $M$ varies between gases, $c_p – c_v$ is different for different gases, whereas molar $C_p – C_v = R$ is universal!
3. Heat Capacities in Terms of Degrees of Freedom ($f$)
Combining the equipartition theorem with thermodynamics expresses all caloric parameters as direct functions of active quadratic degrees of freedom:
$$\mathbf{C_v = \frac{f}{2} R}, \qquad \mathbf{C_p = \left(\frac{f}{2} + 1\right) R = \frac{f + 2}{2} R}, \qquad \mathbf{\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}}$$
| Gas Molecular Type | Degrees of Freedom ($f$) | $C_v$ | $C_p = C_v + R$ | $\gamma = C_p / C_v$ | $C_p – C_v$ |
|---|---|---|---|---|---|
| Monoatomic ($\text{He}, \text{Ne}, \text{Ar}$) | $3$ ($3$ trans) | $\frac{3}{2} R$ | $\frac{5}{2} R$ | $\frac{5}{3} \approx 1.67$ | $R$ |
| Diatomic (Rigid) ($\text{H}_2, \text{N}_2, \text{O}_2$) | $5$ ($3$ trans $+ 2$ rot) | $\frac{5}{2} R$ | $\frac{7}{2} R$ | $\frac{7}{5} = 1.40$ | $R$ |
| Diatomic (with Vibration) ($\text{Cl}_2$, High $T$) | $7$ ($3$ trans $+ 2$ rot $+ 2$ vib) | $\frac{7}{2} R$ | $\frac{9}{2} R$ | $\frac{9}{7} \approx 1.29$ | $R$ |
| Linear Polyatomic ($\text{CO}_2, \text{C}_2\text{H}_2$) | $5$ (rigid room $T$) | $\frac{5}{2} R$ | $\frac{7}{2} R$ | $\frac{7}{5} = 1.40$ | $R$ |
| Non-Linear Polyatomic ($\text{H}_2\text{O}, \text{NH}_3, \text{CH}_4$) | $6$ ($3$ trans $+ 3$ rot) | $3 R$ | $4 R$ | $\frac{4}{3} \approx 1.33$ | $R$ |
4. General Theory of Gas Mixtures
For a non-reacting mixture of ideal gases containing $n_1, n_2, \dots$ moles with respective molar heat capacities $C_{v1}, C_{v2}, \dots$ and adiabatic exponents $\gamma_1, \gamma_2, \dots$:
A. Effective Molar Heat Capacities
By the conservation and additivity of internal energy ($U_{\text{mix}} = \sum U_i$):
$$\mathbf{C_{v,\text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2} + \dots}{n_1 + n_2 + \dots} = \frac{\sum n_i C_{vi}}{\sum n_i}}$$
$$\mathbf{C_{p,\text{mix}} = C_{v,\text{mix}} + R = \frac{\sum n_i C_{pi}}{\sum n_i}}$$
B. Effective Adiabatic Exponent ($\gamma_{\text{mix}}$)
$$\mathbf{\gamma_{\text{mix}} = \frac{C_{p,\text{mix}}}{C_{v,\text{mix}}} = \frac{n_1 C_{p1} + n_2 C_{p2} + \dots}{n_1 C_{v1} + n_2 C_{v2} + \dots}}$$
Expressed directly in terms of individual $\gamma_i$ (since $C_{vi} = \frac{R}{\gamma_i – 1}$):
$$\mathbf{\frac{n_1 + n_2}{\gamma_{\text{mix}} – 1} = \frac{n_1}{\gamma_1 – 1} + \frac{n_2}{\gamma_2 – 1}}$$
C. Effective Degrees of Freedom
$$\mathbf{f_{\text{mix}} = \frac{\sum n_i f_i}{\sum n_i} \implies \gamma_{\text{mix}} = 1 + \frac{2}{f_{\text{mix}}}}$$
5. Critical JEE Trap Alerts
1. Universal Internal Energy Equation:
$\Delta U = n C_v \Delta T$ applies to ANY ideal gas process (isochoric, isobaric, isothermal, adiabatic, or polytropic), because $U$ is a state function depending solely on temperature!
2. Molar vs. Mass Basis:
Always verify whether gas ratios are given by mass ($m_i$) or moles ($n_i$). If given by mass, first calculate moles via $n_i = m_i / M_i$ before applying mixture heat capacity formulas.
3. Isobaric Heat Fractionation:
For any isobaric expansion of an ideal gas:
– Fraction of heat converting to internal energy: $\frac{\Delta U}{Q} = \frac{C_v}{C_p} = \frac{1}{\gamma}$.
– Fraction of heat converting to work: $\frac{W}{Q} = \frac{R}{C_p} = 1 – \frac{1}{\gamma} = \frac{\gamma – 1}{\gamma}$.
Four Graded Solved Examples
Example 1 (JEE Main Level: Mayer’s Relation & Specific Heat Capacities)
Problem Statement: For an unknown ideal gas, the difference between molar heat capacities is $C_p – C_v = 8.314\text{ J/(mol}\cdot\text{K)}$ and the ratio of heat capacities is measured to be $\gamma = 1.40 = 7/5$.
1. Calculate the values of $C_v$ and $C_p$ in $\text{J/(mol}\cdot\text{K)}$.
2. Identify the molecular structure and atomicity of the gas molecules at room temperature.
3. If this gas is pure Nitrogen ($\text{N}_2$, molar mass $M = 28.0\text{ g/mol}$), determine the specific heat capacities per unit mass ($c_v$ and $c_p$) in $\text{J/(kg}\cdot\text{K)}$.
Solution:
Step 1: Calculate $C_v$ and $C_p$.
From $\gamma = \frac{C_p}{C_v} \implies C_p = \gamma C_v$.
Substituting into Mayer’s relation $C_p – C_v = R$:
$$\gamma C_v – C_v = R \implies C_v (\gamma – 1) = R$$
$$C_v = \frac{R}{\gamma – 1} = \frac{8.314\text{ J/(mol}\cdot\text{K)}}{1.40 – 1.0} = \frac{8.314}{0.40} = \mathbf{20.785\text{ J/(mol}\cdot\text{K)}} \quad \left(=\frac{5}{2} R\right)$$
$$C_p = \gamma C_v = 1.40 \times 20.785\text{ J/(mol}\cdot\text{K)} = \mathbf{29.099\text{ J/(mol}\cdot\text{K)}} \quad \left(=\frac{7}{2} R\right)$$
Step 2: Identify Molecular Structure.
From $C_v = \frac{f}{2} R = \frac{5}{2} R$, the number of active degrees of freedom is $f = 5$. This unambiguously identifies the gas as a rigid diatomic gas (or linear polyatomic gas) with $3$ translational and $2$ rotational degrees of freedom.
Step 3: Specific Heat Capacities per Unit Mass.
Molar mass of $\text{N}_2$: $M = 28.0\text{ g/mol} = 0.0280\text{ kg/mol}$.
$$c_v = \frac{C_v}{M} = \frac{20.785\text{ J/(mol}\cdot\text{K)}}{0.0280\text{ kg/mol}} \approx \mathbf{742.3\text{ J/(kg}\cdot\text{K)}}$$
$$c_p = \frac{C_p}{M} = \frac{29.099\text{ J/(mol}\cdot\text{K)}}{0.0280\text{ kg/mol}} \approx \mathbf{1039.3\text{ J/(kg}\cdot\text{K)}}$$
Verification of Specific Mayer’s Relation:
$$c_p – c_v = 1039.25 – 742.32 = 296.93\text{ J/(kg}\cdot\text{K)} = \frac{8.314}{0.0280}\text{ J/(kg}\cdot\text{K)} \quad (\text{Exact!})$$
Example 2 (Mathematical Formulation: Mass-Based Gas Mixture & Effective Quantities)
Problem Statement: A rigid vessel contains a gas mixture consisting of $m_1 = 4.0\text{ g}$ of Helium ($\text{He}$, monoatomic, $M_1 = 4.0\text{ g/mol}$) and $m_2 = 14.0\text{ g}$ of Nitrogen ($\text{N}_2$, rigid diatomic, $M_2 = 28.0\text{ g/mol}$). Take $R = 8.314\text{ J/(mol}\cdot\text{K)}$.
1. Determine the effective molar mass $M_{\text{mix}}$ of the mixture in $\text{g/mol}$.
2. Find the effective molar heat capacities $C_{v,\text{mix}}$ and $C_{p,\text{mix}}$ of the mixture expressed in terms of $R$.
3. Find the effective adiabatic index $\gamma_{\text{mix}}$ of the mixture as a ratio of co-prime integers.
4. Compute the specific heat capacity at constant volume of the mixture ($c_{v,\text{mix}}$) in $\text{J/(kg}\cdot\text{K)}$.
Solution:
Step 1: Mole Quantities & Effective Molar Mass.
$$n_{\text{He}} = \frac{m_1}{M_1} = \frac{4.0\text{ g}}{4.0\text{ g/mol}} = 1.0\text{ mol}$$
$$n_{\text{N}_2} = \frac{m_2}{M_2} = \frac{14.0\text{ g}}{28.0\text{ g/mol}} = 0.50\text{ mol}$$
Total moles: $n_{\text{total}} = 1.0 + 0.50 = 1.50\text{ mol}$.
Total mass: $m_{\text{total}} = 4.0 + 14.0 = 18.0\text{ g}$.
$$M_{\text{mix}} = \frac{m_{\text{total}}}{n_{\text{total}}} = \frac{18.0\text{ g}}{1.50\text{ mol}} = \mathbf{12.0\text{ g/mol}} = 0.0120\text{ kg/mol}$$
Step 2: Effective Molar Heat Capacities.
For Helium (monoatomic): $C_{v1} = \frac{3}{2} R$. For Nitrogen (diatomic): $C_{v2} = \frac{5}{2} R$.
$$C_{v,\text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{(1.0 \times \frac{3}{2} R) + (0.50 \times \frac{5}{2} R)}{1.50} = \frac{1.5 R + 1.25 R}{1.50} = \frac{2.75}{1.50} R = \mathbf{\frac{11}{6} R \approx 1.833 R}$$
$$C_{p,\text{mix}} = C_{v,\text{mix}} + R = \frac{11}{6} R + R = \mathbf{\frac{17}{6} R \approx 2.833 R}$$
Step 3: Effective Adiabatic Index ($\gamma_{\text{mix}}$).
$$\gamma_{\text{mix}} = \frac{C_{p,\text{mix}}}{C_{v,\text{mix}}} = \frac{17/6 R}{11/6 R} = \mathbf{\frac{17}{11} \approx 1.545}$$
Step 4: Specific Heat Capacity ($c_{v,\text{mix}}$).
$$c_{v,\text{mix}} = \frac{C_{v,\text{mix}}}{M_{\text{mix}}} = \frac{\frac{11}{6} \times 8.314\text{ J/(mol}\cdot\text{K)}}{0.0120\text{ kg/mol}} = \frac{15.2423}{0.0120} \approx \mathbf{1270.2\text{ J/(kg}\cdot\text{K)}}$$
Example 3 (JEE Advanced Level: Work, Heat & Energy Fractions Across Paths)
Problem Statement: One mole of a rigid diatomic ideal gas ($C_v = \frac{5}{2} R, C_p = \frac{7}{2} R$) expands from initial state $(P_0, V_0)$ to a final volume of $2V_0$ along two distinct paths:
– Path 1: An isobaric expansion at constant pressure $P = P_0$.
– Path 2: A process in which pressure is directly proportional to volume ($P = \alpha V$).
For each path:
1. Express the heat absorbed ($Q$), work performed ($W$), and change in internal energy ($\Delta U$) in terms of $P_0 V_0$.
2. Determine the fraction of absorbed heat converted into external work ($W/Q$) and the fraction stored as internal energy ($\Delta U/Q$).
Solution:
Analysis of Path 1 (Isobaric Expansion):
– Work done: $W_1 = P_0 (V_f – V_i) = P_0 (2V_0 – V_0) = \mathbf{P_0 V_0}$.
– Temperature change: $\Delta (P V) = P_0 (2V_0) – P_0 V_0 = P_0 V_0 = R \Delta T_1$.
– Internal energy change: $\Delta U_1 = C_v \Delta T_1 = \frac{5}{2} R \Delta T_1 = \mathbf{\frac{5}{2} P_0 V_0}$.
– Heat absorbed: $Q_1 = \Delta U_1 + W_1 = \frac{5}{2} P_0 V_0 + P_0 V_0 = \mathbf{\frac{7}{2} P_0 V_0}$.
– Energy fractions:
$$\frac{W_1}{Q_1} = \frac{P_0 V_0}{\frac{7}{2} P_0 V_0} = \mathbf{\frac{2}{7} \approx 28.6\%}, \qquad \frac{\Delta U_1}{Q_1} = \frac{\frac{5}{2} P_0 V_0}{\frac{7}{2} P_0 V_0} = \mathbf{\frac{5}{7} \approx 71.4\%}$$
Analysis of Path 2 ($P = \alpha V$):
At initial state: $P_0 = \alpha V_0 \implies \alpha = P_0 / V_0$.
At final state: $V_f = 2V_0 \implies P_f = \alpha (2V_0) = 2P_0$.
– Work done (area under the linear $P-V$ curve):
$$W_2 = \frac{P_0 + P_f}{2} (V_f – V_i) = \frac{P_0 + 2P_0}{2} (2V_0 – V_0) = \mathbf{\frac{3}{2} P_0 V_0}$$
– Internal energy change:
$$\Delta (P V) = P_f V_f – P_0 V_0 = (2P_0)(2V_0) – P_0 V_0 = 4P_0 V_0 – P_0 V_0 = 3P_0 V_0 = R \Delta T_2$$
$$\Delta U_2 = C_v \Delta T_2 = \frac{5}{2} R \Delta T_2 = \frac{5}{2} (3P_0 V_0) = \mathbf{\frac{15}{2} P_0 V_0}$$
– Heat absorbed:
$$Q_2 = \Delta U_2 + W_2 = \frac{15}{2} P_0 V_0 + \frac{3}{2} P_0 V_0 = \frac{18}{2} P_0 V_0 = \mathbf{9 P_0 V_0}$$
Alternative via Molar Heat Capacity: Process $P V^{-1} = \text{const} \implies n = -1$.
$$C = C_v + \frac{R}{1 – (-1)} = \frac{5}{2}R + \frac{1}{2}R = 3R \implies Q_2 = n C \Delta T = 3 R \Delta T_2 = 3(3P_0 V_0) = 9P_0 V_0 \quad (\text{Exact!})$$
– Energy fractions:
$$\frac{W_2}{Q_2} = \frac{\frac{3}{2} P_0 V_0}{9 P_0 V_0} = \mathbf{\frac{1}{6} \approx 16.7\%}, \qquad \frac{\Delta U_2}{Q_2} = \frac{\frac{15}{2} P_0 V_0}{9 P_0 V_0} = \mathbf{\frac{5}{6} \approx 83.3\%}$$
Example 4 (JEE Advanced Level: High-Temperature Gas Mixture & Sound Speed)
Problem Statement: An enclosure holds $n_1 = 2.0\text{ moles}$ of Helium ($\text{He}$, monoatomic, $M_1 = 4.0\text{ g/mol}$) and $n_2 = 1.0\text{ mole}$ of Iodine vapor ($\text{I}_2$, $M_2 = 254.0\text{ g/mol}$) at high temperature $T$ where the vibrational mode of $\text{I}_2$ is fully active ($f_2 = 3\text{ trans} + 2\text{ rot} + 2\text{ vib} = 7$).
1. Determine the effective degrees of freedom $f_{\text{mix}}$ and the adiabatic exponent $\gamma_{\text{mix}}$ of the mixture.
2. Calculate the ratio of the speed of sound in this gas mixture to the speed of sound in pure Helium at the same absolute temperature ($v_{s,\text{mix}} / v_{s,\text{He}}$).
Solution:
Step 1: Effective Degrees of Freedom & $\gamma_{\text{mix}}$.
For Helium: $f_1 = 3$. For Iodine vapor at high $T$: $f_2 = 7$.
$$f_{\text{mix}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2} = \frac{(2.0 \times 3) + (1.0 \times 7)}{2.0 + 1.0} = \frac{6 + 7}{3} = \mathbf{\frac{13}{3} \approx 4.333}$$
$$\gamma_{\text{mix}} = 1 + \frac{2}{f_{\text{mix}}} = 1 + \frac{2}{13/3} = 1 + \frac{6}{13} = \mathbf{\frac{19}{13} \approx 1.4615}$$
Step 2: Effective Molar Mass of the Mixture.
$$M_{\text{mix}} = \frac{n_1 M_1 + n_2 M_2}{n_1 + n_2} = \frac{(2.0 \times 4.0) + (1.0 \times 254.0)}{3.0} = \frac{8.0 + 254.0}{3.0} = \frac{262.0}{3.0} \approx \mathbf{87.33\text{ g/mol}}$$
Step 3: Ratio of Speed of Sound.
The speed of sound in an ideal gas is $v_s = \sqrt{\frac{\gamma R T}{M}}$. At identical temperature $T$:
$$\frac{v_{s,\text{mix}}}{v_{s,\text{He}}} = \sqrt{\frac{\gamma_{\text{mix}} / M_{\text{mix}}}{\gamma_{\text{He}} / M_{\text{He}}}} = \sqrt{\frac{\gamma_{\text{mix}}}{\gamma_{\text{He}}} \cdot \frac{M_{\text{He}}}{M_{\text{mix}}}}$$
Substituting known values ($\gamma_{\text{He}} = 5/3$, $M_{\text{He}} = 4.0\text{ g/mol}$):
$$\frac{v_{s,\text{mix}}}{v_{s,\text{He}}} = \sqrt{\frac{19/13}{5/3} \cdot \frac{4.0}{262/3}} = \sqrt{\frac{57}{65} \cdot \frac{12.0}{262.0}} = \sqrt{\frac{684}{17030}} \approx \sqrt{0.040164} \approx \mathbf{0.2004}$$
The speed of sound in the mixture is approximately $20.0\%$ of its speed in pure Helium, primarily because the massive Iodine molecules dramatically increase the effective inertia of the medium.
Practice Worksheet: 10 JEE Problems
Questions 1 to 5: JEE Main Level (Single Correct)
Q1. For an ideal gas obeying the classical ideal gas law, the difference between the molar heat capacities at constant pressure and constant volume ($C_p – C_v$) is:
(A) Directly proportional to the atomicity of the gas
(B) Strictly equal to the universal gas constant $R$, independent of atomicity
(C) Inversely proportional to the molar mass of the gas
(D) Equal to zero for monoatomic gases
Q2. An ideal gas has an adiabatic index $\gamma = 1.33 = 4/3$. Its molar heat capacity at constant volume $C_v$ in terms of the universal gas constant $R$ is:
(A) $\frac{3}{2} R$
(B) $\frac{5}{2} R$
(C) $3 R$
(D) $4 R
Q3. When an ideal rigid diatomic gas undergoes an isobaric expansion, the fraction of the total heat energy absorbed that is converted into external mechanical work is:
(A) $\frac{2}{7}$
(B) $\frac{5}{7}$
(C) $\frac{2}{5}$
(D) $\frac{1}{3}$
Q4. A container contains an equimolar mixture of a monoatomic gas ($\gamma_1 = 5/3$) and a rigid diatomic gas ($\gamma_2 = 7/5$). The effective adiabatic index $\gamma_{\text{mix}}$ of the mixture is:
(A) $1.40$
(B) $1.50$
(C) $1.53$
(D) $1.60$
Q5. Helium gas has a molar mass of $M = 4.0\text{ g/mol}$. Given $R = 8.314\text{ J/(mol}\cdot\text{K)}$, its specific heat capacity at constant volume per unit mass ($c_v$) is approximately:
(A) $2078\text{ J/(kg}\cdot\text{K)}$
(B) $3118\text{ J/(kg}\cdot\text{K)}$
(C) $5196\text{ J/(kg}\cdot\text{K)}$
(D) $1247\text{ J/(kg}\cdot\text{K)}$
Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)
Q6. Which of the following statements regarding the heat capacities of ideal gases is/are correct?
(A) For any ideal gas, $C_p – C_v = R$ holds true regardless of whether the gas is monoatomic, diatomic, or polyatomic.
(B) The ratio $\gamma = C_p / C_v$ monotonically decreases as the number of active molecular degrees of freedom increases.
(C) For specific heat capacities per unit mass, $c_p – c_v = R / M$, where $M$ is the molar mass of the gas.
(D) In an isothermal expansion or compression of an ideal gas, the effective molar heat capacity is infinite ($C \to \infty$).
Q7. A gas mixture contains $n_1$ moles of gas 1 (with parameters $C_{v1}, \gamma_1$) and $n_2$ moles of gas 2 (with parameters $C_{v2}, \gamma_2$). Which of the following expressions correctly describe the mixture?
(A) The effective molar heat capacity at constant volume is $C_{v,\text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}$.
(B) The effective heat capacities satisfy Mayer’s relation: $C_{p,\text{mix}} – C_{v,\text{mix}} = R$.
(C) The effective adiabatic exponent satisfies $\frac{n_1 + n_2}{\gamma_{\text{mix}} – 1} = \frac{n_1}{\gamma_1 – 1} + \frac{n_2}{\gamma_2 – 1}$.
(D) For any arbitrary thermodynamic process with temperature change $\Delta T$, the change in internal energy of the mixture is $\Delta U = (n_1 + n_2) C_{v,\text{mix}} \Delta T$.
Q8. An ideal gas with adiabatic exponent $\gamma$ undergoes a reversible polytropic process along the path $P V^n = \text{constant}$. Which of the following statements is/are correct?
(A) The molar heat capacity during the process is $C = \frac{R}{\gamma – 1} + \frac{R}{1 – n}$.
(B) In the range $1 < n < \gamma$, the molar heat capacity $C$ is negative, meaning temperature increases while heat is extracted.
(C) When $n = \gamma$, the process is adiabatic and $C = 0$.
(D) When $n = 0$, the process is isobaric and $C = C_p$.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. One mole of a rigid diatomic ideal gas ($C_v = \frac{5}{2} R$) undergoes a quasi-static expansion process in which its pressure is directly proportional to its volume ($P = \alpha V$). When the temperature of the gas increases by $\Delta T = 20\text{ K}$, the heat absorbed by the gas is given by $Q = k \cdot R$, where $k$ is an integer. Determine the value of $k$.
Q10. A gas mixture contains $2.0\text{ moles}$ of Helium ($\text{He}$, monoatomic, $\gamma_1 = 5/3$) and $3.0\text{ moles}$ of a rigid diatomic gas ($\gamma_2 = 7/5$). The effective adiabatic index of the mixture is expressed in simplest fractional form as $\gamma_{\text{mix}} = \frac{m}{n}$, where $m$ and $n$ are co-prime positive integers ($\gcd(m, n) = 1$). Find the value of $(m + n)$.
Detailed Solutions & Answer Key
Answer Key
| Question | Correct Option / Answer | Topic / Concept |
|---|---|---|
| Q1 | (B) Strictly equal to $R$ | Mayer’s Relation for Ideal Gases |
| Q2 | (C) $3 R$ | Molar Heat Capacity from $\gamma = 4/3$ |
| Q3 | (A) $\frac{2}{7}$ | Isobaric Work Fraction of Diatomic Gas |
| Q4 | (B) $1.50$ | Adiabatic Exponent of Equimolar Gas Mixture |
| Q5 | (B) $3118\text{ J/(kg}\cdot\text{K)}$ | Specific Heat Capacity of Helium |
| Q6 | (A), (B), (C), (D) | Comprehensive Heat Capacity Properties |
| Q7 | (A), (B), (C), (D) | Gas Mixture Equivalence Relations |
| Q8 | (A), (B), (C), (D) | Polytropic Heat Capacity & Negative Specific Heat |
| Q9 | 60 | Heat Absorbed in $P \propto V$ Process ($C = 3R$) |
| Q10 | 52 | Mixture Exponent $\gamma = 31/21 \implies m + n = 52$ |
Complete Step-by-Step Solutions
Solution 1:
From Mayer’s relation $C_p – C_v = R$, the difference between the molar heat capacities at constant pressure and constant volume for an ideal gas is identically equal to the universal gas constant $R = 8.314\text{ J/(mol}\cdot\text{K)}$, completely independent of molecular atomicity, geometry, or temperature.
Correct Option: (B)
Solution 2:
From $\gamma = \frac{C_p}{C_v} = \frac{C_v + R}{C_v} = 1 + \frac{R}{C_v}$:
$$\gamma – 1 = \frac{R}{C_v} \implies C_v = \frac{R}{\gamma – 1}$$
Given $\gamma = 4/3$:
$$C_v = \frac{R}{4/3 – 1} = \frac{R}{1/3} = \mathbf{3 R}$$
Correct Option: (C)
Solution 3:
For any isobaric expansion of an ideal gas:
$$W = P \Delta V = n R \Delta T, \qquad Q = n C_p \Delta T$$
The fraction of heat converted to work is:
$$\frac{W}{Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p}$$
For a rigid diatomic gas, $C_p = \frac{7}{2} R$:
$$\frac{W}{Q} = \frac{R}{\frac{7}{2} R} = \mathbf{\frac{2}{7}}$$
Correct Option: (A)
Solution 4:
For equal moles ($n_1 = n_2 = n$):
$$f_1 = 3 \quad (\text{monoatomic}), \qquad f_2 = 5 \quad (\text{rigid diatomic})$$
$$f_{\text{mix}} = \frac{n(3) + n(5)}{2n} = \frac{8}{2} = 4$$
$$\gamma_{\text{mix}} = 1 + \frac{2}{f_{\text{mix}}} = 1 + \frac{2}{4} = 1 + 0.50 = \mathbf{1.50}$$
Correct Option: (B)
Solution 5:
For Helium (monoatomic, $M = 4.0\text{ g/mol} = 0.0040\text{ kg/mol}$):
$$C_v = \frac{3}{2} R = 1.5 \times 8.314\text{ J/(mol}\cdot\text{K)} = 12.471\text{ J/(mol}\cdot\text{K)}$$
The specific heat capacity per unit mass is:
$$c_v = \frac{C_v}{M} = \frac{12.471\text{ J/(mol}\cdot\text{K)}}{0.0040\text{ kg/mol}} \approx \mathbf{3117.75\text{ J/(kg}\cdot\text{K)} \approx 3118\text{ J/(kg}\cdot\text{K)}}$$
Correct Option: (B)
Solution 6:
(A) Correct: $C_p – C_v = R$ is universally valid for any ideal gas.
(B) Correct: $\gamma = 1 + 2/f$. As $f$ increases, $2/f$ decreases, so $\gamma$ decreases.
(C) Correct: Dividing $C_p – C_v = R$ by molar mass $M$ yields $c_p – c_v = R/M$.
(D) Correct: For an isothermal process, $dT = 0$ while heat is exchanged ($dQ \neq 0$), so $C = dQ / (n dT) \to \infty$.
Correct Options: (A), (B), (C), (D)
Solution 7:
(A) Correct: By conservation and additivity of internal energy.
(B) Correct: $C_{p,\text{mix}} – C_{v,\text{mix}} = \frac{\sum n_i (C_{pi} – C_{vi})}{\sum n_i} = \frac{\sum n_i R}{\sum n_i} = R$.
(C) Correct: Since $C_{vi} = \frac{R}{\gamma_i – 1}$, substituting into $C_{v,\text{mix}} = \frac{\sum n_i C_{vi}}{\sum n_i}$ directly proves this standard mixture relation.
(D) Correct: Internal energy is a state function: $\Delta U = n_{\text{total}} C_{v,\text{mix}} \Delta T$.
Correct Options: (A), (B), (C), (D)
Solution 8:
(A) Correct: For $P V^n = \text{const}$, $W = \frac{n R \Delta T}{1 – n}$, giving $C = C_v + \frac{R}{1 – n} = \frac{R}{\gamma – 1} + \frac{R}{1 – n}$.
(B) Correct: In the interval $1 < n < \gamma$, the second term $-\frac{R}{n - 1}$ dominates the positive term $\frac{R}{\gamma - 1}$, resulting in $C < 0$.
(C) Correct: For $n = \gamma$, $C = \frac{R}{\gamma – 1} – \frac{R}{\gamma – 1} = 0$, representing an adiabatic process.
(D) Correct: For $n = 0$, $P = \text{const} \implies C = C_v + R = C_p$.
Correct Options: (A), (B), (C), (D)
Solution 9:
For the process $P \propto V \implies P V^{-1} = \text{constant}$.
This is a polytropic process with polytropic exponent $n = -1$.
The molar heat capacity of the gas is:
$$C = C_v + \frac{R}{1 – n} = \frac{5}{2} R + \frac{R}{1 – (-1)} = \frac{5}{2} R + \frac{1}{2} R = \mathbf{3 R}$$
The total heat absorbed by $n_{\text{moles}} = 1.0\text{ mol}$ during temperature change $\Delta T = 20\text{ K}$ is:
$$Q = n_{\text{moles}} C \Delta T = 1.0 \times (3 R) \times 20 = \mathbf{60 R}$$
Thus, $k = 60$.
Answer: 60
Solution 10:
For Helium ($n_1 = 2.0$, monoatomic): $f_1 = 3$.
For the rigid diatomic gas ($n_2 = 3.0$): $f_2 = 5$.
The effective degrees of freedom of the mixture:
$$f_{\text{mix}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2} = \frac{(2.0 \times 3) + (3.0 \times 5)}{2.0 + 3.0} = \frac{6 + 15}{5} = \frac{21}{5} = \mathbf{4.2}$$
The adiabatic exponent of the mixture is:
$$\gamma_{\text{mix}} = 1 + \frac{2}{f_{\text{mix}}} = 1 + \frac{2}{21/5} = 1 + \frac{10}{21} = \mathbf{\frac{31}{21}}$$
Here $m = 31$ and $n = 21$, which are co-prime positive integers ($\gcd(31, 21) = 1$).
$$m + n = 31 + 21 = \mathbf{52}$$
Answer: 52