Thermodynamics in Physics Questions and Answers

Answer: C

The source must be kept at 327 °C.

With temperatures in kelvin, the efficiency is \(\eta = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{300}{400} = 0.25\). Doubling it to 0.5 with the sink still at 300 K:

\[ \begin{aligned} 1 - \frac{300}{T_{1}'} &= 0.5 \\ &\Rightarrow T_{1}' \\ &= 600\ \text{K} \\ &= 327^{\circ}\text{C} \end{aligned} \]

600 °C is the trap of forgetting to convert back from kelvin, and the other temperatures do not give an efficiency of 0.5.

Answer: A

The largest percentage change is in its volume.

If the linear expansion coefficient is \(\alpha\), a small temperature rise \(\Delta T\) changes every length (radius, diameter) by a fraction \(\alpha\Delta T\). Area depends on length squared and volume on length cubed, so for small changes:

\[ \begin{aligned} \frac{\Delta r}{r} &= \alpha\Delta T, \quad \frac{\Delta A}{A} = 2\alpha\Delta T \\ \frac{\Delta V}{V} &= 3\alpha\Delta T \end{aligned} \]

Volume changes by three times the percentage of the radius, more than any other option. Radius and diameter change by the same percentage, the smallest. Mass does not change at all on heating.

Answer: E

Global warming leads to the expansion of desert.

A warmer climate increases evaporation and shifts rainfall patterns, so many dry regions become drier and deserts spread (desertification), one of the recognised consequences of global warming.

Each of the other options states the opposite of what happens. Global warming raises, not lowers, the average temperature of the earth; it melts ice caps faster and warms the oceans, so the sea level rises rather than falls; and it makes weather more variable and extreme rather than static.

Answer: A

The relation is \(C_{v} = \frac{R}{\Upsilon - 1}\), where the paper writes \(\Upsilon\) for the ratio of specific heats usually called \(\gamma\).

For one mole of an ideal gas, \(C_{p} - C_{v} = R\) and \(\frac{C_{p}}{C_{v}} = \gamma\). Substituting \(C_{p} = \gamma C_{v}\):

\[ \begin{aligned} \gamma C_{v} - C_{v} &= R \\ C_{v} &= \frac{R}{\gamma - 1} \end{aligned} \]

Then \(C_{p} = \frac{\gamma R}{\gamma - 1}\). \(\frac{R}{1 - \gamma}\) and \(\frac{R\gamma}{1 - \gamma}\) are negative because \(\gamma \gt 1\), so they cannot be heat capacities. \(\frac{R}{1 + \gamma}\) has the wrong sign in the denominator.

Answer: E

Both its temperature and internal energy increase.

An insulated vessel allows no heat exchange, so the compression is adiabatic: \(Q = 0\). By the first law, \(\Delta U = Q + W_{\text{on}} = W_{\text{on}}\). Compressing the gas does positive work on it, so its internal energy rises, and for an ideal gas internal energy depends only on temperature, so the temperature rises too.

Internal energy and temperature therefore cannot decrease. In a compression the volume falls while the pressure rises, so 'both pressure and volume increase' and 'both temperature and volume increase' are both wrong.

Answer: C

It is called a diathermic wall.

In the zeroth law of thermodynamics, a diathermic wall is a conducting wall that lets heat pass between two systems, so that they come to the same temperature, thermal equilibrium.

An adiabatic wall is the opposite: an ideal insulator that allows no heat flow, so 'insulated wall' and 'non-conducting wall' describe the adiabatic case too. 'Semiconducting wall' is not a thermodynamic term.

Answer: E

The mismatch is Adiabatic : Heat exchange.

An adiabatic process is defined by no heat exchange with the surroundings (\(q = 0\)); any change in internal energy comes from work alone.

The other pairs are as the paper intends. Isobaric means constant pressure, and isochoric constant volume. In an isothermal process the temperature is fixed, so heat must be absorbed or emitted to balance the work done. An irreversible process is associated with dissipative losses, such as friction, which the paper describes as loss of heat.

Answer: A

The efficiency is 75 %.

The efficiency of a heat engine is the fraction of absorbed heat converted to work:

\[ \begin{aligned} \eta &= 1 - \frac{Q_{2}}{Q_{1}} \\ &= 1 - \frac{1}{4} \\ &= \frac{3}{4} \\ &= 75\% \end{aligned} \]

Here \(Q_{2}\) is the heat rejected to the sink and \(Q_{1}\) the heat absorbed from the source. 25 % is the fraction rejected, not converted. 50 % and 60 % do not follow from a 1 : 4 ratio.

Answer: B

The ratio is 0.6.

For a Carnot engine:

\[ \begin{aligned} \eta &= 1 - \frac{T_{2}}{T_{1}} = 0.4 \\ \frac{T_{2}}{T_{1}} &= 1 - 0.4 = 0.6 \end{aligned} \]

0.4 is the efficiency itself, not the temperature ratio.

So if the source is at 500 K, the sink is at 300 K. The ratio is always \(1 - \eta\), whichever temperatures are chosen.

Answer: A

The work done is \(Q_{1} - Q_{2}\).

By the first law, \(Q = \Delta U + W\). At constant temperature (for an ideal gas) or over a complete cycle, \(\Delta U = 0\), so the work done equals the net heat taken in: heat supplied \(Q_{1}\) minus heat given out \(Q_{2}\).

\(Q_{1} + Q_{2}\) adds the heat expelled instead of subtracting it, and the halved forms have no basis in the first law.

This is the same balance used for a heat engine over a cycle, where the efficiency is \(\frac{Q_{1} - Q_{2}}{Q_{1}}\).