Two moles of helium gas (gamma = 5/3) are initially at temperature 27°C and occupy a volume of 20 L.

Q: Why does the temperature double during the isobaric expansion?

According to Charles's Law, at constant pressure, the volume of an ideal gas is directly proportional to its absolute temperature ($V \propto T$). Since the volume doubles, the temperature must also double.

Q: Why do we use $T^\gamma p^{1-\gamma} = \text{constant}$ instead of $pV^\gamma = \text{constant}$?

We use $T^\gamma p^{1-\gamma} = \text{constant}$ because we know the initial and final temperatures of the adiabatic process, and we want to find the final pressure. It's a direct relation that saves us from calculating the volume first.

Q: How do we calculate the work done in an adiabatic process?

In an adiabatic process, there is no heat exchange ($Q = 0$). According to the first law of thermodynamics ($\Delta U = Q - W$), the work done by the gas equals the negative change in its internal energy ($W = -\Delta U = nC_V(T_{\text{initial}} - T_{\text{final}})$).

Q: Why is $C_V = \frac{3}{2}R$ for helium?

Helium is a monoatomic gas. Monoatomic gases have 3 translational degrees of freedom and no rotational or vibrational degrees of freedom. Therefore, their molar heat capacity at constant volume is $C_V = \frac{f}{2}R = \frac{3}{2}R$.

Q: Can we find the work done by calculating the area under the $p$-$V$ curve?

Yes! The work done is exactly the area under the $p$-$V$ curve. For the isobaric part, it's a simple rectangle. For the adiabatic part, you would integrate $p \, dV$, which mathematically results in the same formula we used: $W = \frac{p_1 V_1 - p_2 V_2}{\gamma - 1}$.

Question

Two moles of helium gas ($\gamma = 5/3$) are initially at temperature $27^{\circ}	ext{C}$ and occupy a volume of $20	ext{ L}$. The gas is first expanded at constant pressure until the volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value.
(a) Sketch the process on a $p$-$V$ diagram.
(b) What are the final volume and pressure of the gas?
(c) What is the work done by the gas?

Sigma Percentile · Accepted Answer

The problem takes us on a thermodynamic journey with two moles of helium gas. We are tasked with tracking its state through an isobaric expansion followed by an adiabatic expansion, and finally calculating the total work done. Let's break this down step-by-step.

### Analyzing the Setup

We start with $n = 2$ moles of helium. Since helium is a monoatomic gas, its adiabatic index is $\gamma = 5/3$. The initial state, let's call it State A, has a temperature $T_A = 27^\circ	ext{C} = 300	ext{ K}$ and a volume $V_A = 20	ext{ L} = 20 	imes 10^{-3}	ext{ m}^3$.

To fully define State A, we need its pressure. We can find this using the **ideal gas law**:

$$p_A = \frac{nRT_A}{V_A}$$

Substituting the known values:

$$p_A = \frac{2 	imes 8.31 	imes 300}{20 	imes 10^{-3}} = 2.49 	imes 10^5	ext{ N/m}^2$$

### The Isobaric Expansion

The gas first expands at a constant pressure until its volume doubles. This is an **isobaric process** from State A to State B.

Because the pressure is constant, $p_B = p_A = 2.49 	imes 10^5	ext{ N/m}^2$. The volume doubles, so $V_B = 40	ext{ L}$. According to **Charles's Law**, at constant pressure, volume is directly proportional to temperature ($V \propto T$). Therefore, the temperature also doubles:

$$T_B = 2T_A = 600	ext{ K}$$

### The Adiabatic Expansion

Next, the gas undergoes an **adiabatic expansion** until its temperature returns to the initial value. This takes us to State C, where $T_C = 300	ext{ K}$.

To find the final pressure $p_C$, we use the adiabatic relation between temperature and pressure:

$$T^\gamma p^{1-\gamma} = 	ext{constant}$$

Rearranging this to solve for $p_C$:

$$p_C = p_B \left(\frac{T_B}{T_C}ight)^{\frac{\gamma}{1-\gamma}}$$

For helium, $\frac{\gamma}{1-\gamma} = \frac{5/3}{1 - 5/3} = -\frac{5}{2}$. Substituting the values:

$$p_C = (2.49 	imes 10^5) \left(\frac{600}{300}ight)^{-5/2} = \frac{2.49 	imes 10^5}{2^{2.5}} \approx 0.44 	imes 10^5	ext{ N/m}^2$$

Similarly, to find the final volume $V_C$, we use the relation between temperature and volume:

$$T V^{\gamma-1} = 	ext{constant}$$

Rearranging for $V_C$:

$$V_C = V_B \left(\frac{T_B}{T_C}ight)^{\frac{1}{\gamma-1}}$$

Substituting the values:

$$V_C = 40 	imes \left(\frac{600}{300}ight)^{3/2} = 40 	imes 2^{1.5} \approx 113	ext{ L}$$

### Calculating the Work Done

The total work done by the gas is the sum of the work done in each process.

For the isobaric process A to B, the work done is the constant pressure times the change in volume:

$$W_{AB} = p_A (V_B - V_A)$$

$$W_{AB} = (2.49 	imes 10^5) 	imes (40 - 20) 	imes 10^{-3} = 4980	ext{ J}$$

For the adiabatic process B to C, there is no heat exchange ($Q = 0$). By the **first law of thermodynamics**, the work done equals the negative change in internal energy:

$$W_{BC} = -\Delta U = nC_V(T_B - T_C)$$

For a monoatomic gas, $C_V = \frac{3}{2}R$. Substituting the values:

$$W_{BC} = 2 	imes \left(\frac{3}{2} 	imes 8.31ight) 	imes (600 - 300) = 7479	ext{ J}$$

Finally, the net work done is the sum of these two values:

$$W_{	ext{net}} = W_{AB} + W_{BC} = 4980 + 7479 = 12459	ext{ J}$$

This completes our comprehensive analysis of the thermodynamic cycle!

Visualized Solution

$n = 2 moles, γ = 5/3$

$p_{A} = \frac{n R T _{A}}{V _{A}}$

$p_{B} = p_{A} = 2.49 \times 1 0^{5} N/m^{2}$

$T_{C} = T_{A} = 300 K$

$p_{C} = 0.44 \times 1 0^{5} N/m^{2}$

$V_{C} = 113 L$

$W_{A B} = p_{A} (V_{B} - V_{A})$

$W_{B C} = - Δ U = n C_{V} (T_{B} - T_{C})$

$W_{net} = W_{A B} + W_{B C}$

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup

The Isobaric Expansion

The Adiabatic Expansion

Calculating the Work Done

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=2 moles,γ=5/3

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } pA​=VA​nRTA​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } pB​=pA​=2.49×105 N/m2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TC​=TA​=300 K

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } pC​=0.44×105 N/m2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VC​=113 L

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } WAB​=pA​(VB​−VA​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } WBC​=−ΔU=nCV​(TB​−TC​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Wnet​=WAB​+WBC​