Thermodynamics MCQs with Answers – Part 5 (Class 11 Physics)

Explanation: A Carnot refrigerator is a reversible ideal device between two reservoirs. It gives the maximum possible coefficient of performance for those reservoir temperatures. Real refrigerators have irreversibilities such as friction, electrical losses, and heat transfer through finite temperature differences. These effects require extra work input for the same cooling effect. A COP greater than \(1\) is still possible, but it will not exceed the Carnot limit for the same temperatures.

Explanation: Free expansion into vacuum is not a quasistatic process. The gas does not pass through a well-defined sequence of equilibrium states during the expansion. Since the vessel is insulated, \(Q=0\). Since the gas expands into vacuum, the external pressure is zero and the external work is \(W=0\). The first law then gives \(\Delta U=0\), and for an ideal gas this means no temperature change. The isothermal and reversible adiabatic path formulas are not appropriate for the intermediate free-expansion path.

Explanation: For an ideal gas, \(\Delta U\) depends only on temperature. If the initial and final temperatures are the same, both processes have \(\Delta U=0\). In a quasistatic isothermal expansion, the gas can do work and absorb heat with \(Q=W\). In insulated free expansion into vacuum, \(Q=0\) and external work is \(W=0\). The same internal-energy change does not imply the same heat, work, or path nature.

Explanation: \( \textbf{First-law relation:} \) \[ Q=\Delta U+W \] \( \textbf{Solve for \(\Delta U\):} \) \[ \Delta U=Q-W \] \( \textbf{Substitute heat and work:} \) \[ \Delta U=500-700=-200\,\text{J} \] \( \textbf{Ideal-gas internal-energy relation:} \) \[ \Delta U=nC_V\Delta T \] \( \textbf{Given combined value:} \) \[ nC_V=50\,\text{J K}^{-1} \] \( \textbf{Temperature change:} \) \[ \Delta T=\frac{-200}{50}=-4\,\text{K} \] \( \textbf{Physical reading:} \) The gas does more work than the heat supplied, so its internal energy and temperature decrease. \( \textbf{Final answer:} \) \(\Delta T=-4\,\text{K}\).

Explanation: Row P correctly pairs the logarithmic work formula with an isothermal quasistatic process of a fixed ideal gas. Row Q is wrong because \(PV^\gamma=\text{constant}\) requires a reversible adiabatic ideal-gas process. Row R is wrong because Carnot formulas require absolute temperatures in \(\text{K}\). Row S is wrong because \(W=P\Delta V\) with one value of \(P\) applies to constant-pressure work; when pressure changes, the area or integral form is needed. A formula is safe only when its process condition is satisfied.

Explanation: Heat-engine efficiency compares useful work output with heat absorbed from the hot reservoir, so \(\eta=\frac{W}{Q_1}\). Refrigerator coefficient of performance compares useful heat removed from the cold region with work input, so \(\beta=\frac{Q_2'}{W'}\). Both are ratios of energies, so both are dimensionless. Reversing either ratio changes its physical meaning. Multiplying energy terms would not give a performance measure.

Explanation: \( \textbf{Cycle condition:} \) Over a complete cycle, \(\Delta U_{\text{cycle}}=0\). \( \textbf{Heat absorbed:} \) \(+1000\,\text{J}\). \( \textbf{Heat rejected:} \) \(-400\,\text{J}\). \( \textbf{Net heat:} \) \[ Q_{\text{net}}=1000-400=600\,\text{J} \] \( \textbf{First law over a cycle:} \) \[ Q_{\text{net}}=W_{\text{net}} \] \( \textbf{Net work delivered:} \) \[ W_{\text{net}}=600\,\text{J} \] \( \textbf{Compression note:} \) Individual step work may be negative, but the net cycle work is fixed by net heat because \(\Delta U_{\text{cycle}}=0\). \( \textbf{Final answer:} \) The net work delivered is \(600\,\text{J}\).

Explanation: \( \textbf{Carnot efficiency:} \) \[ \eta_C=1-\frac{T_2}{T_1} \] \( \textbf{Substitution:} \) \[ \eta_C=1-\frac{400}{1000}=0.60 \] \( \textbf{Actual efficiency:} \) \(\eta_{\text{actual}}=0.50\). \( \textbf{Comparison:} \) \[ 0.50<0.60 \] \( \textbf{Conclusion:} \) An efficiency below the Carnot limit is physically possible in principle. \( \textbf{Work output check:} \) \[ W=(0.50)(5000)=2500\,\text{J} \] \( \textbf{Final answer:} \) The engine is below the Carnot limit and is not ruled out by it.

Explanation: A smooth \(P\)-\(V\) path assumes that the system has well-defined pressure and volume at each stage. This is possible when the process is quasistatic, meaning the system passes through a sequence of near-equilibrium states. Sudden free expansion and turbulent mixing are not usually represented by a regular system-pressure path. Heat flow through a finite temperature difference may occur between bodies, but it does not by itself guarantee a well-defined gas path. The graph condition is mainly about equilibrium-like intermediate states.

Explanation: Row P is correct because internal energy is a state function and returns to its original value in a complete cycle. Heat and work are path-dependent transfers, so their net values over a cycle may be non-zero. Row Q is wrong because an adiabatic process has \(Q=0\), not necessarily \(\Delta U=0\). Row R is wrong because isothermal ideal-gas expansion can have \(Q=W\). Row S is wrong because a heat engine must reject some heat to a sink when operating cyclically between finite reservoirs.

Explanation: For an ideal gas, internal energy depends only on temperature. Since both paths start from the same state and end at the same final temperature, the change in internal energy is fixed. Heat supplied and work done are path-dependent, so they can differ for the two routes. The area under the \(P\)-\(V\) curve represents work, and different paths can give different areas. The same final state fixes \(\Delta U\), not the separate values of \(Q\) and \(W\).

Explanation: \( \textbf{Given data:} \) \(n=1.0\,\text{mol}\), \(T_i=300\,\text{K}\), \(T_f=500\,\text{K}\), \(R=8.0\,\text{J mol}^{-1}\text{K}^{-1}\), and \(C_V=20\,\text{J mol}^{-1}\text{K}^{-1}\). \( \textbf{Temperature rise:} \) \[ \Delta T=500-300=200\,\text{K} \] \( \textbf{Find \(C_P\) from Mayer’s relation:} \) \[ C_P=C_V+R \] \[ C_P=20+8=28\,\text{J mol}^{-1}\text{K}^{-1} \] \( \textbf{Constant-pressure heat relation:} \) \[ Q_P=nC_P\Delta T \] \( \textbf{Substitution:} \) \[ Q_P=(1.0)(28)(200)\,\text{J} \] \( \textbf{Calculation:} \) \[ Q_P=5600\,\text{J} \] \( \textbf{Energy meaning:} \) This heat includes both internal-energy increase and expansion work. \( \textbf{Final answer:} \) \(Q_P=5600\,\text{J}\).

Explanation: For an ideal gas, \(\Delta U=nC_V\Delta T\) gives the internal-energy change. At constant pressure, heating usually makes the gas expand. Expansion means the gas does pressure-volume work on the surroundings. Therefore, heat supplied at constant pressure is \(Q_P=\Delta U+W\), not just \(\Delta U\). This is why \(C_P\) is greater than \(C_V\) for gases.

Explanation: In isothermal expansion, the gas temperature remains equal to its initial temperature. In adiabatic expansion, no heat is supplied while the gas does work. The work output comes from internal energy, so the ideal gas cools. At the same final volume, the adiabatic path has lower temperature and therefore lower pressure than the isothermal path. This comparison is a process-condition result, not just a volume-change result.

Explanation: \( \textbf{Given data:} \) \(T_i=480\,\text{K}\), \(V_f=3V_i\), and \(\gamma=\frac{4}{3}\). \( \textbf{Adiabatic temperature-volume relation:} \) \[ TV^{\gamma-1}=\text{constant} \] \( \textbf{Apply between initial and final states:} \) \[ T_iV_i^{\gamma-1}=T_fV_f^{\gamma-1} \] \( \textbf{Solve for final temperature:} \) \[ T_f=T_i\left(\frac{V_i}{V_f}\right)^{\gamma-1} \] \( \textbf{Exponent:} \) \[ \gamma-1=\frac{4}{3}-1=\frac{1}{3} \] \( \textbf{Substitution:} \) \[ T_f=480\left(\frac{V_i}{3V_i}\right)^{1/3} \] \[ T_f=480\left(\frac{1}{3}\right)^{1/3} \] \( \textbf{Cube-root evaluation:} \) \[ 3^{1/3}\approx1.44 \] \[ T_f\approx\frac{480}{1.44}\approx333\,\text{K} \] \( \textbf{Physical check:} \) The gas expands adiabatically, so its temperature must fall below \(480\,\text{K}\), but not to zero. \( \textbf{Final answer:} \) The final temperature is closest to \(333\,\text{K}\).

Explanation: \( \textbf{Given data:} \) \(T_i=600\,\text{K}\), \(\frac{V_f}{V_i}=8\), and \(\gamma=\frac{4}{3}\). \( \textbf{Adiabatic temperature-volume relation:} \) \[ TV^{\gamma-1}=\text{constant} \] \( \textbf{Solve for final temperature:} \) \[ T_f=T_i\left(\frac{V_i}{V_f}\right)^{\gamma-1} \] \( \textbf{Exponent value:} \) \[ \gamma-1=\frac{4}{3}-1=\frac{1}{3} \] \( \textbf{Substitution:} \) \[ T_f=600\left(\frac{1}{8}\right)^{1/3} \] \( \textbf{Cube-root step:} \) \[ 8^{1/3}=2 \] \[ \left(\frac{1}{8}\right)^{1/3}=\frac{1}{2} \] \( \textbf{Calculation:} \) \[ T_f=600\times\frac{1}{2}=300\,\text{K} \] \( \textbf{Final answer:} \) The final temperature is \(300\,\text{K}\).

Explanation: For an isothermal ideal-gas process, the work done by the gas is \(W=nRT\ln\left(\frac{V_f}{V_i}\right)\). Here \(V_i=4V\) and \(V_f=V\). Therefore, \(\frac{V_f}{V_i}=\frac{1}{4}\). Since \(\ln\left(\frac{1}{4}\right)=-\ln4\), the work is \(W=-nRT\ln4\). The negative sign shows that work is done on the gas during compression.

Explanation: Pressure-volume work requires volume change. In an isochoric process, the volume is constant, so \(dV=0\). Therefore, \(W=\int P\,dV=0\) for each isochoric part. In the isobaric expansion, the volume changes at constant pressure, so work is non-zero and positive for work done by the gas. The graph area comes only from the part where the path moves horizontally in volume.

Explanation: \( \textbf{Cycle shape:} \) The cycle is rectangular on a \(P\)-\(V\) graph. \( \textbf{Pressure height:} \) \[ \Delta P=4P_0-P_0=3P_0 \] \( \textbf{Volume width:} \) \[ \Delta V=3V_0-V_0=2V_0 \] \( \textbf{Area enclosed:} \) \[ \text{Area}=\Delta P\,\Delta V \] \[ \text{Area}=(3P_0)(2V_0)=6P_0V_0 \] \( \textbf{Sign rule:} \) A clockwise cycle gives positive net work done by the gas. \( \textbf{Final answer:} \) \(W_{\text{net}}=+6P_0V_0\).

Explanation: \( \textbf{Initial hot temperature:} \) \(T_{1i}=600\,\text{K}\). \( \textbf{Final hot temperature:} \) \(T_{1f}=900\,\text{K}\). \( \textbf{Cold temperature:} \) \(T_2=300\,\text{K}\). \( \textbf{Initial Carnot efficiency:} \) \[ \eta_i=1-\frac{300}{600}=1-\frac{1}{2}=\frac{1}{2} \] \( \textbf{Final Carnot efficiency:} \) \[ \eta_f=1-\frac{300}{900}=1-\frac{1}{3}=\frac{2}{3} \] \( \textbf{Increase:} \) \[ \eta_f-\eta_i=\frac{2}{3}-\frac{1}{2} \] \[ \eta_f-\eta_i=\frac{4-3}{6}=\frac{1}{6} \] \( \textbf{Final answer:} \) The increase in efficiency is \(\frac{1}{6}\).