Thermodynamics in Physics Questions and Answers

Answer: B

The heat absorbed is 6 kJ.

Convert to kelvin: \(T_{1} = 600\) K and \(T_{2} = 500\) K.

\[ \begin{aligned} \eta &= 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{500}{600} = \frac{1}{6} \\ Q_{1} &= \frac{W}{\eta} = 1 \times 6 = 6\ \text{kJ} \end{aligned} \]

Using Celsius temperatures by mistake gives \(\eta = 1 - \frac{227}{327} \approx 0.31\) and about 3.3 kJ, close to the 3 kJ distractor. Carnot efficiency always needs absolute temperatures.

Answer: D

The pressure becomes 32P.

A sudden compression is adiabatic, so \(PV^{\gamma}\) is constant, with \(\gamma = \frac{5}{3}\) for a monoatomic gas.

\[ \begin{aligned} P' &= P\left(\frac{V}{V/8}\right)^{5/3} = P \cdot 8^{5/3} \\ &= P \cdot (2^{3})^{5/3} = 2^{5}P = 32P \end{aligned} \]

The option 8P is the isothermal (Boyle's law) answer, which would apply to a slow compression at constant temperature, not a sudden one.

Answer: E

The heat required is 200 \(C_{V}\).

The question works per mole: \(C_{V}\) here is the molar heat capacity at constant volume. Heat supplied at constant volume is

\[ \begin{aligned} Q &= n C_{V} \Delta T \\ &= 2 \times C_{V} \times (127 - 27) = 200\,C_{V} \end{aligned} \]

A temperature difference is the same in degrees Celsius and kelvin, so there is no need to convert to 300 K and 400 K. 100 \(C_{V}\) forgets the 2 moles, and 50 \(C_{V}\) halves instead of doubling; 300 \(C_{V}\) and 500 \(C_{V}\) do not follow from a change of 100 degrees.

Answer: E

The heat required is 796 kJ.

Three stages: warm the ice to 0°C, melt it, then warm the water to 10°C.

\[ \begin{aligned} Q_{1} &= 2 \times 2100 \times 10 = 42\ \text{kJ} \\ Q_{2} &= 2 \times 3.35 \times 10^{5} = 670\ \text{kJ} \\ Q_{3} &= 2 \times 4200 \times 10 = 84\ \text{kJ} \\ Q &= 42 + 670 + 84 = 796\ \text{kJ} \end{aligned} \]

335 kJ is the latent heat for 1 kg only, and 398 kJ is the whole sum for 1 kg instead of 2. The melting stage dominates, so leaving it out (126 kJ) is far off every option.

Answer: C

The efficiency changes by 25%.

For a heat engine, \(\eta = 1 - \frac{Q_{2}}{Q_{1}}\), where \(Q_{2}\) is heat rejected and \(Q_{1}\) heat absorbed.

\[ \begin{aligned} \eta_{1} &= 1 - \tfrac{1}{4} = 75\% \\ \eta_{2} &= 1 - \tfrac{1}{2} = 50\% \\ \Delta\eta &= 25\% \end{aligned} \]

75% and 50% are the two efficiencies themselves; the question asks for the change, a fall of 25 percentage points.

Answer: D

The final pressure is \(100\sqrt{2}\) Pa.

'Suddenly compressed' means adiabatic, so \(PV^{\gamma}\) is constant:

\[ \begin{aligned} P_{2} &= P_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma} = 50 \times 2^{1.5} \\ &= 50 \times 2\sqrt{2} = 100\sqrt{2} \end{aligned} \]

100 Pa is the isothermal answer (Boyle's law, \(\gamma\) taken as 1), and \(50\sqrt{2}\) uses the power 0.5.

Answer: A

There is no change in internal energy in an isothermal process.

The internal energy of an ideal gas depends only on its temperature, \(U = nC_V T\). In an isothermal process \(T\) is constant, so \(\Delta U = 0\) and all the heat supplied goes into work, \(Q = W\).

In an adiabatic process \(Q = 0\) but \(\Delta U = -W \neq 0\), so the temperature changes. In isobaric and isochoric processes the temperature changes too, so \(\Delta U \neq 0\).

Answer: D

The variable that is not extensive is density.

An extensive variable scales with the size of the system: double the system and it doubles. Total mass, internal energy and volume all do this, and so does the work \(P\,dV\), since it is proportional to the volume change.

Density is mass divided by volume. Doubling the system doubles both, so density is unchanged: it is an intensive variable, like pressure and temperature.

Answer: E

The work done is 600 J.

The efficiency of a Carnot engine depends only on the two temperatures.

\[ \begin{aligned} \eta &= 1 - \frac{T_2}{T_1} = 1 - \frac{300}{600} = 0.5 \\ W &= \eta Q_1 = 0.5 \times 1200 = 600\ \text{J} \end{aligned} \]

1200 J would mean 100% efficiency, which no heat engine can reach, and 2400 J or more would be more work than the heat taken in. The other 600 J is rejected to the sink at 300 K.

Answer: A

The thermal stress is \(1 \times 10^{8}\) Nm^-2.

If the rod is prevented from expanding, the strain it is denied is \(\alpha\Delta T\), and the stress is Young's modulus times that strain:

\[ \begin{aligned} \text{stress} &= Y\alpha\Delta T \\ &= 10^{11} \times 2 \times 10^{-5} \times 50 \\ &= 10^{8}\ \text{Nm}^{-2} \end{aligned} \]

The length of 1 m does not matter, since stress depends on strain, not on the actual expansion. \(2 \times 10^{8}\) is twice the correct value, and the \(10^{6}\) values lose two powers of ten.