First Law of Thermodynamics

Consider one mole of an ideal gas.

1. Heating at constant volume. Let heat $dQ_V$ raise its temperature by $dT$. Then $dQ_V = C_V\,dT$. Since $dV = 0$, no work is done, and by the first law

$$dQ_V = dU \;\Rightarrow\; dU = C_V\,dT \qquad \text{(i)}$$

The internal energy of an ideal gas depends only on its temperature, so (i) gives $dU$ for the same $dT$ in any process. 2. Heating at constant pressure. Let heat $dQ_P$ raise the temperature by the same $dT$, while the volume increases by $dV$. Then $dQ_P = C_P\,dT$, and by the first law

$$dQ_P = dU + P\,dV \;\Rightarrow\; C_P\,dT = C_V\,dT + P\,dV \qquad \text{(ii)}$$

3. Using the gas equation. For one mole, $PV = RT$. At constant pressure, differentiating gives

$$P\,dV = R\,dT \qquad \text{(iii)}$$

4. Substituting (iii) into (ii):

$$C_P\,dT = C_V\,dT + R\,dT$$

5. Dividing by $dT$:

$$\boxed{C_P - C_V = R}$$

This is Mayer's relation. $C_P$ is greater than $C_V$ by $R$, which is the extra heat per mole per kelvin needed to do the work of expansion at constant pressure.

(d) For cyclic process, $\Delta W = 0$

In a cyclic process the system returns to its initial state, so the change in internal energy is zero ($\Delta U = 0$). The net work done equals the area enclosed by the cycle on the P–V diagram, which is generally not zero. The other three statements are correct.

(a) Adiabatic process

An adiabatic process is a thermodynamic process in which no heat enters or leaves the system ($dQ = 0$). Pressure, volume and temperature may all change. It occurs in a perfectly insulated container, or when the change is so sudden that there is no time for heat exchange. For an ideal gas, $PV^\gamma = \text{constant}$.

(b) Work done in an adiabatic process

1. Let $n$ moles of an ideal gas expand adiabatically from state $(P_1, V_1, T_1)$ to $(P_2, V_2, T_2)$.
2. Work done by the gas when the volume changes by $dV$ is $dW = P\,dV$, so

$$W = \int_{V_1}^{V_2} P\,dV$$

3. For an adiabatic process, $PV^\gamma = K$ (constant), so $P = KV^{-\gamma}$:

$$W = K\int_{V_1}^{V_2} V^{-\gamma}\,dV = K\left[\frac{V^{1-\gamma}}{1-\gamma}\right]_{V_1}^{V_2} = \frac{KV_2^{1-\gamma} - KV_1^{1-\gamma}}{1-\gamma}$$

4. Since $K = P_1V_1^\gamma = P_2V_2^\gamma$, we have $KV_2^{1-\gamma} = P_2V_2$ and $KV_1^{1-\gamma} = P_1V_1$:

$$W = \frac{P_2V_2 - P_1V_1}{1-\gamma}$$

5. Hence

$$\boxed{W = \frac{P_1V_1 - P_2V_2}{\gamma - 1}}$$

6. Using $PV = nRT$, this can also be written as

$$\boxed{W = \frac{nR(T_1 - T_2)}{\gamma - 1}}$$

In an adiabatic expansion $T_2 < T_1$, so $W$ is positive: the gas does work at the cost of its internal energy and cools down.

(c) 259 K

$T_1 = 72 + 273 = 345$ K, $V_2 = 8V_1$, $\gamma - 1 = \tfrac23$. From $TV^{\gamma-1} = \text{constant}$:

$$T_2 = T_1\left(\frac{V_1}{V_2}\right)^{2/3} = 345\times\left(\frac18\right)^{2/3} = \frac{345}{4} \approx 86\ \text{K}$$

Change $= 345 - 86 \approx 259$ K. Note that an adiabatic expansion gives a fall in temperature of about 259 K, not a rise. ($86$ K is the final temperature, not the change.)

By the first law, $dQ = dU + P\,dV$.

(i) At constant volume: $dV = 0$, so no external work is done ($dW = P\,dV = 0$). All the heat added goes into increasing the internal energy of the gas, and so raises its temperature:

$$dQ_V = dU = nC_V\,dT$$

(ii) At constant pressure: the gas expands while it is heated. Part of the heat increases the internal energy (raising the temperature) and the rest is used to do external work $P\,dV$ in expanding against the surroundings:

$$dQ_P = dU + P\,dV = nC_P\,dT$$

So, to raise the temperature by the same amount, more heat is needed at constant pressure. This is why $C_P > C_V$.

(b) $5P_1V_1$

(The question means an ideal gas.) The net work in a cycle equals the area enclosed on the P–V diagram. Here the cycle is a right triangle with base $3V_1 - V_1 = 2V_1$ and height $6P_1 - P_1 = 5P_1$:

$$W = \tfrac12\times2V_1\times5P_1 = 5P_1V_1$$

The cycle is traced clockwise, so this work is done by the gas (positive).

Given:

• Initial temperature, $T_1 = 30^\circ$C $= 30 + 273 = 303$ K
• Final volume, $V_2 = \dfrac{8}{27}V_1$
• $\gamma = \dfrac53$, so $\gamma - 1 = \dfrac23$

A sudden compression is adiabatic, so

$$T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1} \;\Rightarrow\; T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}$$

Substituting:

$$T_2 = 303\times\left(\frac{27}{8}\right)^{2/3} = 303\times\left[\left(\frac32\right)^3\right]^{2/3} = 303\times\left(\frac32\right)^2 = 303\times\frac94$$ $$T_2 = 681.75\ \text{K}$$

Change in temperature:

$$\Delta T = T_2 - T_1 = 681.75 - 303$$ $$\boxed{\Delta T \approx 379\ \text{K}\ (\text{a rise of about } 379^\circ\text{C})}$$

The temperature rises because the work done on the gas during compression increases its internal energy.

(A) $\dfrac45$

At constant volume $P \propto T$ (in kelvin). $T_1 = 127 + 273 = 400$ K and $T_2 = 227 + 273 = 500$ K, so

$$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{400}{500} = \frac45$$

(In the figure, the upper pressure level is read as $4P_1$.)

a) Indicator diagram

An indicator diagram is a graph of pressure ($P$) against volume ($V$) of a working substance, showing how its state changes during a thermodynamic process or cycle. The area under the curve gives the work done, and for a cycle the area enclosed by the loop gives the net work done.

b) Work done in process II

Process II is at constant pressure $4P_1$, with the volume increasing from $V_1$ to $3V_1$:

$$W_{II} = P\,\Delta V = 4P_1\times(3V_1 - V_1) = 4P_1\times2V_1$$ $$\boxed{W_{II} = 8P_1V_1}$$

This work is done by the gas, because it expands.

c) Identifying the processes

(D) The system is insulated from the surroundings

In an adiabatic process no heat is exchanged ($dQ = 0$), which happens when the system is thermally insulated (or the change is very fast). Its pressure, volume and temperature can all change.

First law of thermodynamics: when a quantity of heat $dQ$ is supplied to a system, part of it increases the internal energy of the system ($dU$) and the rest is used to do external work ($dW$):

$$dQ = dU + dW = dU + P\,dV$$

Yes, it follows the principle of conservation of energy. The first law is simply the law of conservation of energy applied to heat and work. The heat given to a system is neither created nor destroyed; it is fully accounted for as the increase in internal energy plus the work done by the system. For example, a gas cannot do work without either receiving heat or losing an equal amount of internal energy, so a machine that gives work without any energy input (a perpetual motion machine of the first kind) is impossible.

(d) Isochoric

In an isochoric process the volume does not change ($\Delta V = 0$), so the work done $W = \int P\,dV = 0$. This is the minimum possible.

(c) $6P_1V_1$

The net work done by the gas in a cycle equals the area enclosed by the loop on the $P$-$V$ diagram. Reading the enclosed pressure and volume spans from the figure, the area of the closed loop is $6P_1V_1$. The work is positive because the cycle is traversed clockwise.

(a) First law of thermodynamics

The heat supplied to a system equals the sum of the increase in its internal energy and the external work done by it: $Q = \Delta U + W$.

(b) Relation between $C_p$ and $C_v$

For one mole heated through $dT$ at constant volume, no work is done, so $dQ = C_v\,dT = dU$.

At constant pressure, $dQ = C_p\,dT = dU + P\,dV$. The internal energy of an ideal gas depends only on temperature, so $dU$ is the same. Using $PV = RT$, i.e. $P\,dV = R\,dT$:

$$C_p\,dT = C_v\,dT + R\,dT \;\Rightarrow\; C_p - C_v = R$$

(c) Efficiency of a heat engine

The engine takes heat $Q_1$ from the source at $T_1$, converts part of it to work $W$ and rejects $Q_2$ to the sink at $T_2$. In a complete cycle $\Delta U = 0$, so $W = Q_1 - Q_2$. The efficiency is

$$\eta = \frac{W}{Q_1} = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1}$$

For a reversible (Carnot) engine, $\dfrac{Q_2}{Q_1} = \dfrac{T_2}{T_1}$, so

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

(d) $200\pi$ J

The work done in a cycle equals the area enclosed on the $P$-$V$ diagram. The enclosed region is the half-ellipse under the arc ABC and above the line $P=30$. Its semi-axes are $20\ \text{m}^3$ along $V$ and $20\ \text{N/m}^2$ along $P$ (the figure's axes have different scales), so

$W = \tfrac12\pi (20)(20) = 200\pi\ \text{J}$

The loop is traversed clockwise (ABC on top, return along the bottom), so the work is positive.

(b) Isobaric

By the first law, $Q = \Delta U + W$ with $Q>0$, $\Delta U>0$ and $W>0$. Isochoric has $W=0$, isothermal has $\Delta U=0$, and adiabatic has $Q=0$, so all three are ruled out. Heat that both raises the internal energy and does work is the isobaric case, where the gas expands at constant pressure as it is heated.

(a) The heat supplied to a system equals the sum of the increase in its internal energy and the work done by the system: $Q = \Delta U + W$.

(b) At constant volume, no work is done ($W=0$), so all the heat goes into raising the internal energy: $Q = \Delta U = nC_v\Delta T$. At constant pressure, part of the heat is used to do work in expanding the gas, so only the rest raises the internal energy: $Q = \Delta U + P\Delta V$. For the same $Q$, a smaller part goes into raising the temperature, so the rise in temperature is smaller at constant pressure. Equivalently, $C_p>C_v$, so $\Delta T = Q/(nC)$ is larger for $C_v$.

(c) The work done by the system equals the area under the curve on the $P$-$V$ diagram, $W = \int P\,dV$. The work is greatest along the path that lies highest (the largest area under the curve) and least along the path that lies lowest (the smallest area), because all three paths join the same two states and have the same $\Delta V$.

(C) temperature only

In an ideal gas there are no intermolecular forces, so there is no potential energy between the molecules. The internal energy is just the kinetic energy of the molecules, which depends only on the temperature ($U = nC_VT$).

(B) Isobaric

For an expansion from the same initial state to the same final volume, the work done is the area under the P–V curve. In the isobaric process the pressure stays at its initial (highest) value. In the isothermal process it falls, and in the adiabatic process it falls even faster. In the isochoric process the work is zero. So the area, and hence the work, is greatest for the isobaric process: $W_{\text{isobaric}} > W_{\text{isothermal}} > W_{\text{adiabatic}}$.

(c) $ACBDA$

In a cyclic process the net work done equals the area enclosed by the closed loop on the P–V diagram. That is the area $ACBDA$ (the work done along ACB minus the work done along BDA).