Thermodynamics — Practice Questions with Answers

The first law of thermodynamics states that the heat supplied to a system equals the sum of the increase in its internal energy and the work done by the system on its surroundings — a statement of energy conservation.

$\Delta Q = \Delta U + \Delta W$

where $\Delta Q$ is heat supplied, $\Delta U$ is the change in internal energy, and $\Delta W$ is the work done by the system.

Isothermal process: Occurs at constant temperature ($\Delta T = 0$), so internal energy remains constant ($\Delta U=0$ for an ideal gas), and all heat supplied converts entirely to work: $\Delta Q = \Delta W$. Follows $PV = \text{constant}$.

Adiabatic process: Occurs with no heat exchange ($\Delta Q = 0$), so work done is entirely at the expense of internal energy: $\Delta W = -\Delta U$. Follows $PV^\gamma = \text{constant}$, where $\gamma = C_p/C_v$.

Work done: $W = \displaystyle\int_{V_1}^{V_2} P\,dV$. For an ideal gas, $P = \dfrac{nRT}{V}$, and $T$ is constant:

$W = \displaystyle\int_{V_1}^{V_2} \dfrac{nRT}{V}\,dV = nRT\ln\left(\dfrac{V_2}{V_1}\right)$

This can also be written using pressures (since $P_1V_1=P_2V_2$): $W = nRT\ln\left(\dfrac{P_1}{P_2}\right)$

Kelvin-Planck statement: It is impossible to construct a heat engine that, operating in a cycle, converts heat completely into work with no other effect (i.e., without rejecting some heat to a cold reservoir).

This directly implies that a 100% efficient engine is impossible — any real heat engine must always reject some fraction of the heat it absorbs to a cold sink; only a fraction of absorbed heat can be converted to useful work, capping efficiency below 100%.

A Carnot engine is an idealised, reversible heat engine operating in a specific four-step cycle (two isothermal and two adiabatic processes) between two heat reservoirs at fixed temperatures — it represents the theoretical maximum possible efficiency for any heat engine operating between those two temperatures.

$\eta_{Carnot} = 1 - \dfrac{T_2}{T_1}$, where $T_1$ is the temperature of the hot (source) reservoir and $T_2$ is the temperature of the cold (sink) reservoir, both measured in kelvin.

A quasi-static process is one carried out infinitely slowly, such that the system remains arbitrarily close to thermodynamic equilibrium at every intermediate stage throughout the process.

This is an idealisation because real processes always occur at a finite rate, meaning the system is never perfectly in equilibrium throughout. However, treating a process as quasi-static allows the use of equilibrium thermodynamic relations (like the ideal gas law) at every step, greatly simplifying analysis — real slow processes closely approximate this ideal.

Isobaric process: Occurs at constant pressure. Work done = $P\Delta V$ (non-zero, since volume changes).

Isochoric process: Occurs at constant volume. Work done = 0 (since $\Delta V = 0$, and $W = P\Delta V$), so all supplied heat directly increases internal energy: $\Delta Q = \Delta U$.

The Carnot engine represents the theoretical maximum possible efficiency achievable between two given temperatures, since it operates entirely reversibly, with no dissipative losses.

Real engines involve inherently irreversible processes — friction, turbulence, finite-rate heat transfer (not infinitely slow), and other dissipative effects — all of which convert some useful energy into unusable forms (like heat lost to the environment through non-ideal pathways). These unavoidable real-world losses mean actual engines can never achieve the ideal Carnot efficiency for the same operating temperatures.

Internal energy is the total energy contained within a system due to the microscopic kinetic and potential energies of its constituent molecules (translational, rotational, vibrational, and intermolecular potential energy).

It is called a state function because its value depends only on the current state of the system (defined by variables like temperature, pressure, volume) — not on the specific path or process by which that state was reached. This means the change in internal energy between two states is always the same, regardless of how the system got from one state to the other.

Entropy is a measure of disorder or the number of accessible microscopic states of a system — the second law states that the entropy of an isolated system never decreases for any spontaneous process.

When heat flows from a hot body to a cold body, the entropy decrease of the hot body (losing heat) is always smaller in magnitude than the entropy increase of the cold body (gaining the same heat, but at a lower temperature, since $\Delta S = Q/T$) — so total entropy increases, consistent with the second law. The reverse process (cold to hot) would require total entropy to decrease, which is forbidden for a spontaneous, isolated process — this is precisely why heat naturally flows only from hot to cold.

Using the first law: $\Delta Q = \Delta U + \Delta W$

$500 = \Delta U + 200$

$\Delta U = 500 - 200 = \mathbf{300\text{ J}}$

The internal energy of the gas increases by 300 J.

(a) Efficiency: $\eta = 1 - \dfrac{T_2}{T_1} = 1 - \dfrac{300}{600} = 1 - 0.5 = 0.5 = 50\%$

(b) Work done: $W = \eta \times Q_1 = 0.5 \times 800 = 400\text{ J}$

(c) Heat rejected: $Q_2 = Q_1 - W = 800 - 400 = \mathbf{400\text{ J}}$

$W = nRT\ln\left(\dfrac{V_2}{V_1}\right)$

$W = (1)(8.31)(300)\ln\left(\dfrac{20}{10}\right) = 2493\ln(2)$

$\ln(2) \approx 0.693$

$W = 2493 \times 0.693 \approx \mathbf{1728\text{ J}}$

(a) COP: $\text{COP} = \dfrac{Q_2}{W} = \dfrac{100}{25} = \mathbf{4}$

(b) Heat rejected: By energy conservation, $Q_1 = Q_2 + W = 100 + 25 = \mathbf{125\text{ J}}$

Work done (isobaric): $W = P\Delta V = (2\times10^5)(5-2) = 2\times10^5\times3 = 6\times10^5\text{ J}$

Using the first law: $\Delta Q = \Delta U + \Delta W$

$9\times10^5 = \Delta U + 6\times10^5$

$\Delta U = 9\times10^5 - 6\times10^5 = \mathbf{3\times10^5\text{ J}}$

$\eta = 1 - \dfrac{T_2}{T_1}$

$0.6 = 1 - \dfrac{300}{T_1}$

$\dfrac{300}{T_1} = 1 - 0.6 = 0.4$

$T_1 = \dfrac{300}{0.4} = \mathbf{750\text{ K}}$

For a complete cycle, $\Delta U = 0$ (internal energy is a state function, returning to its initial value).

Net heat absorbed: $\Delta Q_{net} = 600 - 350 = 250\text{ J}$

From the first law over the full cycle: $\Delta Q_{net} = \Delta U + \Delta W = 0 + \Delta W$

$\Delta W = \mathbf{250\text{ J}}$

For an isochoric process, $W = 0$, so $\Delta Q = \Delta U = nC_v\Delta T$

$\Delta U = n\left(\dfrac{3}{2}R\right)\Delta T = 2\times\dfrac{3}{2}\times8.31\times(500-300)$

$\Delta U = 3\times8.31\times200 = 24.93\times200$

$\Delta U = \mathbf{4986\text{ J}}$; since $W=0$, heat absorbed $\Delta Q = \Delta U = \mathbf{4986\text{ J}}$ as well.

(a) By energy conservation: $Q_1 = W + Q_2 = 300 + 700 = \mathbf{1000\text{ J}}$

(b) Efficiency: $\eta = \dfrac{W}{Q_1} = \dfrac{300}{1000} = 0.3 = \mathbf{30\%}$

Work done in isothermal expansion is greater than in adiabatic expansion for the same volume change from the same initial state.

Physical explanation: In isothermal expansion, temperature is maintained constant by continuously absorbing heat from a reservoir — this absorbed heat supplies additional energy that helps sustain higher pressure throughout the expansion (since $P = nRT/V$ with $T$ constant), allowing more work to be done ($W=\int P\,dV$).

In adiabatic expansion, no heat is supplied — the gas must do work entirely at the expense of its own internal energy, causing its temperature (and hence pressure, at any given volume) to drop below what it would be in the isothermal case. Since pressure is consistently lower throughout an adiabatic expansion for the same volume change, the resulting work done ($\int P\,dV$) is correspondingly smaller than in the isothermal case.