Kinetic Theory of Gases MCQs with Answers – Part 4 (NEET Physics)

Explanation: \(\textbf{Partition the seven equipartition terms:}\) A rigid diatomic molecule with one active vibration has \[ 3+2+2=7 \] quadratic energy terms. Three are translational, so \[ E_{\rm tr} = \frac37U \] Substituting, \[ E_{\rm tr} = \frac37(2800) = 1200\,\text{J} \] The active vibration contributes two of the seven terms: \[ E_{\rm vib} = \frac27U \] Thus, \[ E_{\rm vib} = \frac27(2800) = 800\,\text{J} \] \(\textbf{Obtain pressure from the translational part:}\) \[ PV=\frac23E_{\rm tr} \] Therefore, \[ PV = \frac23(1200) = 800\,\text{J} \] Hence, \[ P = \frac{800}{8.0\times10^{-3}} = 1.0\times10^5\,\text{Pa} \] \(\textbf{Final answer:}\) \[ \boxed{(1200\,\text{J},800\,\text{J},1.0\times10^5\,\text{Pa})} \]

Explanation: For an ideal gas with \(f_d\) active quadratic energy terms, the equipartition theorem gives \[ U=\frac{f_d}{2}Nk_BT. \] Its total translational kinetic energy is \[ E_{\rm tr}=\frac32Nk_BT. \] Taking the ratio, \[ \frac{U}{E_{\rm tr}} = \frac{f_d}{3}. \] The measured relation gives \[ \frac{U}{E_{\rm tr}}=3. \] Therefore, \[ \frac{f_d}{3}=3, \] which implies \[ f_d=9. \] \(\textbf{Check the allowed molecular modes:}\) In the stated general polyatomic model, the molecule has three translational and three rotational quadratic energy terms. Each active vibrational mode contributes two additional quadratic terms, one kinetic and one potential. Therefore, \[ f_d=6+2r, \] where \(r\) is a non-negative integer representing the number of active vibrational modes. Substituting \(f_d=9\), \[ 9=6+2r \] Hence \[ 2r=3 \] and \[ r=\frac32. \] A molecule cannot have a fractional number of active vibrational modes within this model. Therefore, the measured energy ratio is incompatible with the stated general polyatomic equipartition model. \(\textbf{Final answer:}\) Option A.

Explanation: A monatomic ideal gas has three translational degrees of freedom. Equipartition therefore gives the molar internal energy \[ U=\frac32RT \] for one mole. The molar specific heat at constant volume is \[ C_V=\frac{dU}{dT}. \] Hence, \[ C_V=\frac32R. \] For any ideal gas, \[ C_P-C_V=R. \] Therefore, \[ C_P=\frac32R+R=\frac52R. \] The specific-heat ratio is \[ \gamma=\frac{C_P}{C_V}. \] Thus, \[ \gamma = \frac{5/2}{3/2} = \frac53. \] \(\textbf{Final answer:}\) \[ C_V=\frac32R,\qquad C_P=\frac52R,\qquad \gamma=\frac53. \]

Explanation: \(\textbf{Compare constant-volume heat capacities:}\) For the monatomic gas, \[ C_{V,m}=\frac32R \] For the rigid diatomic gas, \[ C_{V,d}=\frac52R \] At constant volume, \[ Q=\mu C_V\Delta T \] The supplied heat \(Q\) and mole number \(\mu\) are the same. Therefore, \[ C_{V,m}\Delta T_m = C_{V,d}\Delta T_d \] Hence, \[ \frac{\Delta T_m}{\Delta T_d} = \frac{C_{V,d}}{C_{V,m}} = \frac{5/2}{3/2} = \frac53 \] \(\textbf{Relate the temperature rises to pressure rises:}\) For each rigid vessel, \[ PV=\mu RT \] Thus, \[ \Delta P=\frac{\mu R}{V}\Delta T \] The vessels and mole numbers are identical, so \[ \frac{\Delta P_m}{\Delta P_d} = \frac{\Delta T_m}{\Delta T_d} = \frac53 \] \(\textbf{Final answer:}\) \[ \boxed{\left(\frac53,\frac53\right)} \]

Explanation: \(\textbf{Extract the molar constant-volume heat capacity:}\) At constant volume, \[ Q_V=\mu C_V\Delta T \] Therefore, \[ C_V = \frac{Q_V}{\mu\Delta T} \] Substituting, \[ C_V = \frac{4.15\times10^3} {(2.0)(100)} \] Hence, \[ C_V=20.75\,\text{J mol}^{-1}\text{K}^{-1} \] With \(R=8.3\,\text{J mol}^{-1}\text{K}^{-1}\), \[ C_V=\frac52R \] This corresponds to five active quadratic degrees. With no active vibration, the five-degree model is the rigid diatomic gas. For an ideal gas, \[ C_P=C_V+R=\frac72R \] Therefore, \[ \gamma = \frac{C_P}{C_V} = \frac{7/2}{5/2} = \frac75 \] \(\textbf{Final answer:}\) rigid diatomic gas with \[ \gamma=\frac75 \]

Explanation: One active vibrational mode contributes two additional quadratic energy terms. Its molar internal-energy contribution is \[ RT. \] Therefore its contribution to the molar constant-volume heat capacity is \[ R. \] Thus Statement I is correct. For any ideal gas, \[ C_P-C_V=R. \] If \(C_V\) rises by \(R\), then \(C_P=C_V+R\) also rises by \(R\). Therefore Statement II is correct. However, the difference remains \[ C_P-C_V=R, \] so Statement III is incorrect. If initially \[ C_P=C_V+R, \] then \[ \gamma=1+\frac{R}{C_V}. \] Increasing \(C_V\) makes the added fraction \(R/C_V\) smaller. Hence \(\gamma\) decreases, so Statement IV is correct. Therefore I, II and IV are correct.

Explanation: \(\textbf{Interpret the constant-volume heating graph:}\) At constant volume, \[ Q_V=\mu C_V\Delta T \] For equal mole numbers, the graph slope is proportional to \(C_V\). For monatomic X, \[ C_{V,X}=\frac32R \] The slope of Y is twice as large, so \[ C_{V,Y}=2C_{V,X} \] Hence, \[ C_{V,Y}=3R \] \(\textbf{Identify the active-mode structure:}\) For the general polyatomic NCERT model with \(r\) active vibrational modes, \[ C_V=(3+r)R \] Therefore, \[ (3+r)R=3R \] Thus, \[ r=0 \] Then, \[ C_P=C_V+R=4R \] Therefore, \[ \gamma = \frac{C_P}{C_V} = \frac43 \] \(\textbf{Final answer:}\) \[ \boxed{\left(0,\frac43\right)} \]

Explanation: For the rigid diatomic gas with vibration neglected, \[ C_{V,\mathrm{rigid}}=\frac52R \] Activating one vibrational mode adds \[ R \] to \(C_V\). Therefore, \[ C_{V,\mathrm{vib}} = \frac52R+R = \frac72R \] Hence, \[ \frac{C_{V,\mathrm{vib}}}{C_{V,\mathrm{rigid}}} = \frac{7/2}{5/2} = \frac75 \] For the vibrating model, \[ C_P=C_V+R = \frac92R \] Thus, \[ \gamma_{\mathrm{vib}} = \frac{C_P}{C_V} = \frac{9/2}{7/2} = \frac97 \] The additional vibration raises both molar heat capacities and reduces \(\gamma\) from \(7/5\) to \(9/7\). \(\textbf{Final answer:}\) \[ \boxed{\left(\frac75,\frac97\right)} \]

Explanation: For a monatomic ideal gas, \[ C_V=\frac32R, \qquad \gamma=\frac53. \] Therefore, \[ 1\rightarrow p. \] For a rigid diatomic gas, \[ C_V=\frac52R, \qquad \gamma=\frac75, \] so \[ 2\rightarrow q. \] One active vibrational mode adds \(R\) to the rigid-diatomic \(C_V\), giving \[ C_V=\frac72R. \] Then \[ C_P=\frac92R, \qquad \gamma=\frac97, \] so \[ 3\rightarrow r. \] A general polyatomic gas with no active vibration has \[ C_V=3R,\qquad C_P=4R, \] and therefore \[ \gamma=\frac43. \] Thus, \[ 4\rightarrow s. \] Hence the correct matching is \[ 1-p,\;2-q,\;3-r,\;4-s. \]

Explanation: \(\textbf{Infer molar heat capacity from constant-volume heating:}\) \[ Q_V=\mu C_V\Delta T \] Therefore, \[ C_V = \frac{Q_V}{\mu\Delta T} \] Substituting, \[ C_V = \frac{4.15\times10^3} {(0.50)(200)} \] Thus, \[ C_V=41.5\,\text{J mol}^{-1}\text{K}^{-1} \] Using \(R=8.3\,\text{J mol}^{-1}\text{K}^{-1}\), \[ C_V=5R \] \(\textbf{Recover the vibrational-mode count:}\) For the general polyatomic model, \[ C_V=(3+r)R \] Hence, \[ 3+r=5, \] so \[ r=2 \] \(\textbf{Find the constant-pressure heat capacity:}\) For an ideal gas, \[ C_P-C_V=R \] Therefore, \[ C_P=6R \] The constant-pressure heat required is \[ Q_P=\mu C_P\Delta T \] Thus, \[ Q_P = (0.50)(6)(8.3)(200) \] Hence, \[ Q_P=4.98\times10^3\,\text{J} \] \(\textbf{Calculate the specific-heat ratio:}\) \[ \gamma=\frac{C_P}{C_V} = \frac{6R}{5R} = \frac65 \] \(\textbf{Final answer:}\) \[ \boxed{\left(2,\;4.98\times10^3\,\text{J},\;\frac65\right)} \]

Explanation: \(\textbf{Relate heat capacities to the active-degree count:}\) For \(f_d\) active quadratic terms, \[ C_V=\frac{f_d}{2}R. \] For an ideal gas, \[ C_P=C_V+R = \frac{f_d+2}{2}R. \] Therefore, \[ \gamma = \frac{C_P}{C_V} = \frac{f_d+2}{f_d}. \] Using the given value, \[ \frac{f_d+2}{f_d}=\frac97. \] Cross-multiplying, \[ 7f_d+14=9f_d. \] Hence, \[ f_d=7. \] A rigid diatomic molecule contributes five non-vibrational quadratic terms, and one active vibration adds two more: \[ 5+2=7. \] \(\textbf{Final answer:}\) \[ f_d=7, \] corresponding to a rigid diatomic model with one active vibrational mode.

Explanation: For equal mole numbers and equal temperature rises, \[ Q_P=\mu C_P\Delta T \] and \[ Q_V=\mu C_V\Delta T. \] Therefore, \[ \frac{Q_P}{Q_V} = \frac{C_P}{C_V} = \gamma. \] The measured ratio gives \[ \gamma=\frac75. \] For a gas with \(f_d\) active quadratic terms, \[ \gamma=\frac{f_d+2}{f_d}. \] Thus, \[ \frac{f_d+2}{f_d}=\frac75. \] Hence, \[ 5f_d+10=7f_d, \] so \[ f_d=5. \] Five active terms correspond to three translational plus two rotational degrees in the rigid diatomic model with vibration neglected. \(\textbf{Final answer:}\) rigid diatomic ideal gas.

Explanation: For an ideal gas, \[ C_P-C_V=R \] For the same mole number and the same temperature rise, \[ Q_P-Q_V = \mu(C_P-C_V)\Delta T \] Therefore, \[ Q_P-Q_V=\mu R\Delta T \] Substituting the given data, \[ 830=(2.0)R(50) \] Thus, \[ 830=100R \] Hence, \[ R=8.3\,\text{J mol}^{-1}\text{K}^{-1} \] The difference \(Q_P-Q_V\) does not by itself determine whether the gas is monatomic, diatomic or polyatomic. That difference depends on the universal ideal-gas relation \(C_P-C_V=R\), not on the active degree count. \(\textbf{Final answer:}\) \[ \boxed{R=8.3\,\text{J mol}^{-1}\text{K}^{-1}} \]

Explanation: For \(f_d\) active quadratic terms, \[ C_V=\frac{f_d}{2}R. \] Also, \[ C_P=C_V+R. \] Finally, \[ \gamma=\frac{C_P}{C_V}. \] For row B, \[ C_V=\frac52R, \] so \[ C_P=\frac52R+R=\frac72R. \] Therefore, \[ \gamma = \frac{7/2}{5/2} = \frac75. \] All entries in row B are consistent. Row A has the correct heat capacities but the reciprocal value of \(\gamma\). Row C violates \(C_P-C_V=R\). Row D should have \(C_P=5R\), not \(6R\). \(\textbf{Final answer:}\) Row B.

Explanation: \(\textbf{Average the constant-volume capacities over equal mole fractions:}\) For the monatomic gas, \[ C_{V,m}=\frac32R \] For the rigid diatomic gas, \[ C_{V,d}=\frac52R \] Because the mole fractions are equal, \[ C_{V,\mathrm{mix}} = \frac12\left(\frac32R\right) + \frac12\left(\frac52R\right) \] Hence, \[ C_{V,\mathrm{mix}} = 2R \] For an ideal-gas mixture, \[ C_{P,\mathrm{mix}}=C_{V,\mathrm{mix}}+R \] Therefore, \[ C_{P,\mathrm{mix}}=3R \] Thus, \[ \gamma_{\mathrm{mix}} = \frac{3R}{2R} = \frac32 \] \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{C_{V,\mathrm{mix}}}{R},\gamma_{\mathrm{mix}}\right) = \left(2,\frac32\right)} \]

Explanation: \(\textbf{Recover constant-volume heat capacity from the measured heat-capacity ratio:}\) For an ideal gas or ideal-gas mixture, \[ \gamma = 1+\frac{R}{C_V} \] Thus, \[ \frac32 = 1+\frac{R}{C_{V,\mathrm{mix}}} \] Hence, \[ \frac{R}{C_{V,\mathrm{mix}}}=\frac12, \] so \[ C_{V,\mathrm{mix}}=2R \] \(\textbf{Express the mixture heat capacity through composition:}\) The monatomic contribution is \[ \frac32R, \] and the rigid-diatomic contribution is \[ \frac52R \] Therefore, \[ C_{V,\mathrm{mix}} = x\left(\frac32R\right) + (1-x)\left(\frac52R\right) \] Dividing by \(R\), \[ 2=\frac32x+\frac52(1-x) \] Thus, \[ 4=3x+5-5x \] Hence, \[ 2x=1, \qquad x=\frac12 \] \(\textbf{Final answer:}\) \[ \boxed{x=\frac12} \]

Explanation: Equipartition gives \[ C_V=\frac{f_d}{2}R. \] For an ideal gas, \[ C_P=C_V+R. \] Therefore, \[ C_P = \frac{f_d}{2}R+R. \] Combining the terms, \[ C_P = \frac{f_d+2}{2}R. \] Hence, \[ \gamma = \frac{C_P}{C_V} = \frac{(f_d+2)R/2}{f_dR/2}. \] The common factors cancel, giving \[ \gamma=\frac{f_d+2}{f_d}. \] This also shows that increasing the number of active quadratic terms lowers \(\gamma\) toward unity. \(\textbf{Final answer:}\) \[ \gamma=\frac{f_d+2}{f_d}. \]

Explanation: For every ideal gas, \[ C_P-C_V=R \] Therefore, \[ C_P=C_V+R \] The \(C_P\)-versus-\(C_V\) graph is consequently a straight line of unit slope. For the specified point, \[ C_V=4R \] Hence, \[ C_P=4R+R=5R \] The specific-heat ratio is \[ \gamma=\frac{C_P}{C_V} \] Thus, \[ \gamma=\frac{5R}{4R}=\frac54 \] The result is also consistent with \(f_d=8\), because \[ C_V=\frac{f_d}{2}R \] \(\textbf{Final answer:}\) \[ \boxed{(C_P,\gamma)=\left(5R,\frac54\right)} \]

Explanation: \(\textbf{Determine the relevant constant-pressure capacities:}\) For a monatomic ideal gas, \[ C_{V,m}=\frac32R \] Therefore, \[ C_{P,m}=\frac52R \] For the general polyatomic model without active vibration, \[ C_{V,p}=3R \] Hence, \[ C_{P,p}=4R \] At constant pressure, \[ Q=\mu C_P\Delta T \] The heat supplied and mole numbers are equal, so \[ C_{P,m}\Delta T_m = C_{P,p}\Delta T_p \] Therefore, \[ \frac{\Delta T_m}{\Delta T_p} = \frac{C_{P,p}}{C_{P,m}} \] Thus, \[ \frac{\Delta T_m}{\Delta T_p} = \frac{4}{5/2} = \frac85 \] The gas with the smaller constant-pressure heat capacity undergoes the larger temperature rise. \(\textbf{Final answer:}\) \[ \boxed{\frac85} \]

Explanation: \(\textbf{Heat capacity of X at constant volume:}\) With \[ f_X=8, \] equipartition gives \[ C_{V,X} = \frac82R = 4R. \] \(\textbf{Heat capacity of Y at constant pressure:}\) The general polyatomic gas with no active vibration has \[ f_Y=6. \] Thus, \[ C_{V,Y}=3R. \] Therefore, \[ C_{P,Y}=C_{V,Y}+R=4R. \] Hence the relevant heat capacities in the two experiments are equal: \[ C_{V,X}=C_{P,Y}=4R. \] Since \[ Q=\mu C\Delta T \] and both \(Q\) and \(\mu\) are equal, \[ \Delta T_X=\Delta T_Y. \] \(\textbf{Final answer:}\) \[ \Delta T_X=\Delta T_Y. \]