Two moles of an ideal monoatomic gas is taken through a cycle ABCA as shown in the p-T diagram. Duri

Q: Why did we use $pdV = 2nRdT$ for process AB?

Since $pT = \text{constant}$, differentiating gives $pdT + Tdp = 0$. From the ideal gas law $pV = nRT$, differentiating gives $pdV + Vdp = nRdT$. Substituting $V = nRT/p$ and $Tdp = -pdT$, the $Vdp$ term becomes $-nRdT$. Thus, $pdV - nRdT = nRdT$, which simplifies to $pdV = 2nRdT$.

Q: How do we know the gas is monoatomic and what are its heat capacities?

The problem explicitly states it's an 'ideal monoatomic gas'. For monoatomic gases, the degrees of freedom $f = 3$. Therefore, $C_V = \frac{f}{2}R = \frac{3}{2}R$ and $C_p = C_V + R = \frac{5}{2}R$.

Q: Why is the work done in process CA calculated using pressures instead of volumes?

For an isothermal process, Boyle's Law states $p_i V_i = p_f V_f$, which means $V_f/V_i = p_i/p_f$. The standard work formula $nRT \ln(V_f/V_i)$ can thus be rewritten as $nRT \ln(p_i/p_f)$. Both will give the exact same result!

Q: Is the work done in process AB positive or negative?

The work done BY the gas is negative ($-1200R$) because the temperature drops from $600\text{ K}$ to $300\text{ K}$, and since $pV^{1/2} = \text{constant}$, a drop in temperature corresponds to a decrease in volume (compression). Work done ON the gas is positive ($+1200R$).

Q: Why is $\Delta U = 0$ for process CA?

Process CA is a vertical line on the $p-T$ diagram, meaning the temperature is constant ($T = 600\text{ K}$). The internal energy of an ideal gas depends only on its temperature ($U = nC_V T$). Since $\Delta T = 0$, the change in internal energy $\Delta U$ is also zero.

Question

Two moles of an ideal monoatomic gas is taken through a cycle $ABCA$ as shown in the $p$-$T$ diagram. During the process $AB$, pressure and temperature of the gas vary such that $pT = 	ext{constant}$. If $T_1 = 300	ext{ K}$, calculate
(a) the work done on the gas in the process $AB$ and
(b) the heat absorbed or released by the gas in each of the processes.
Give answers in terms of the gas constant $R$.

Sigma Percentile · Accepted Answer

The world of thermodynamics is filled with fascinating cycles, and this problem is a perfect example of how different thermodynamic processes weave together to form a complete story. We are given a $p$-$T$ diagram for two moles of an ideal monoatomic gas undergoing a cyclic process $ABCA$. Our mission? To decode the work done and heat exchanged in each leg of the journey.

Let's take a deep breath and break this down step-by-step.

### Analyzing the Setup

Before we jump into any calculations, we must orient ourselves by identifying the exact coordinates of our states on the $p$-$T$ diagram. The problem gives us $T_1 = 300	ext{ K}$.

Looking at the graph, we can pinpoint the states:
- **State A:** The temperature is $2T_1 = 600	ext{ K}$ and the pressure is $p_1$.
- **State B:** The temperature drops to $T_1 = 300	ext{ K}$ while the pressure rises to $2p_1$.
- **State C:** The temperature goes back up to $2T_1 = 600	ext{ K}$ and the pressure remains at $2p_1$.

We are also told that the gas is monoatomic. This is a crucial piece of intel! It tells us that the molar heat capacity at constant volume is $C_V = \frac{3}{2}R$, and at constant pressure, it is $C_p = \frac{5}{2}R$.

### The Master Equation for Process AB

The path from A to B is not your standard isobaric or isothermal process. We are given a unique constraint: $pT = 	ext{constant}$.

How do we find the work done, $W = \int p dV$, when the relationship is between $p$ and $T$? We need to express $p dV$ entirely in terms of $dT$. Let's use the power of calculus!

Differentiating the given condition $pT = 	ext{constant}$ using the product rule, we get:
$$p dT + T dp = 0 \implies T dp = -p dT$$

Now, let's bring in the trusty ideal gas equation, $pV = nRT$. Differentiating this gives:
$$p dV + V dp = nR dT$$

We need to eliminate the $V dp$ term. From the ideal gas law, we know $V = \frac{nRT}{p}$. Substituting this into our differential equation:
$$p dV + \left(\frac{nRT}{p}ight) dp = nR dT$$

Now, substitute $T dp = -p dT$ into the equation:
$$p dV + \frac{nR}{p} (-p dT) = nR dT$$

$$p dV - nR dT = nR dT$$

Moving the terms around, we arrive at a beautifully simple relation:
$$p dV = 2nR dT$$

### Calculating Work and Heat for Process AB

Now that we have $p dV = 2nR dT$, finding the work done is a breeze. We just integrate from $T_A$ to $T_B$:
$$W_{AB} = \int_{T_A}^{T_B} 2nR dT = 2nR(T_B - T_A)$$

Substitute the known values ($n = 2$, $T_A = 600	ext{ K}$, $T_B = 300	ext{ K}$):
$$W_{AB} = 2(2)R(300 - 600) = -1200 R$$

The negative sign indicates that work is done **on** the gas (compression).

Next, we need the heat exchanged, $Q_{AB}$. According to the First Law of Thermodynamics, $Q = W + \Delta U$. Let's find the change in internal energy:
$$\Delta U_{AB} = nC_V(T_B - T_A) = 2\left(\frac{3}{2}Right)(300 - 600) = -900 R$$

Adding them up:
$$Q_{AB} = -1200 R - 900 R = -2100 R$$

Heat is released by the gas during this process.

### The Isobaric Journey: Process BC

Process B to C is a horizontal line on the $p$-$T$ diagram, which means the pressure is constant. This is an isobaric process!

For an isobaric process, the heat exchanged is directly calculated using $C_p$:
$$Q_{BC} = nC_p(T_C - T_B)$$

Substitute the values ($T_C = 600	ext{ K}$, $T_B = 300	ext{ K}$):
$$Q_{BC} = 2\left(\frac{5}{2}Right)(600 - 300) = 5R(300) = 1500 R$$

Since the result is positive, heat is absorbed by the gas.

### The Isothermal Finale: Process CA

Finally, process C to A is a vertical line on the $p$-$T$ diagram. The temperature is constant at $600	ext{ K}$, making this an isothermal process.

In an isothermal process, the change in internal energy is zero ($\Delta U_{CA} = 0$). Therefore, the heat exchanged is exactly equal to the work done:
$$Q_{CA} = W_{CA} = nRT_C \ln\left(\frac{V_A}{V_C}ight)$$

Since $p_A V_A = p_C V_C$ (Boyle's Law), we can replace the volume ratio with the pressure ratio:
$$Q_{CA} = nRT_C \ln\left(\frac{p_C}{p_A}ight)$$

Substitute the values ($p_C = 2p_1$, $p_A = p_1$):
$$Q_{CA} = 2R(600) \ln\left(\frac{2p_1}{p_1}ight) = 1200 R \ln(2)$$

Given that $\ln(2) \approx 0.693$:
$$Q_{CA} = 1200 R (0.693) = 831.6 R$$

Heat is absorbed by the gas as it expands back to its original state.

### Conclusion

We have successfully navigated the entire cycle! By breaking down the $p$-$T$ diagram and applying the First Law of Thermodynamics to each specific path, we unlocked the secrets of the gas's journey. Remember, the key to mastering thermodynamics is to stay calm, identify the constraints of each process, and trust your fundamental equations. Keep practicing, and these cycles will become second nature!

Visualized Solution

$S t a t e C oor d ina t es$

$P r ocess A B : D i f f er e n t ia l s$

$P r ocess A B : W or k D i f f er e n t ia l$

$P r ocess A B : W or k D o n e$

$P r ocess A B : H e a t E x c han g e d$

$P r ocess B C : I so ba r i c H e a t$

$P r ocess B C : C a l c u l a t i o n$

$P r ocess C A : I so t h er ma l H e a t$

$P r ocess C A : C a l c u l a t i o n$

$C y c l e S u mma r y$

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup

The Master Equation for Process AB

Calculating Work and Heat for Process AB

The Isobaric Journey

The Isothermal Finale

Conclusion

Similar Questions

One mole of a monatomic ideal gas undergoes the cyclic process J $\to$ K $\to$ L $\to$ M $\to$ J, as shown in the P-T diagram. Match the quantities mentioned in List-I with their values in List-II and choose the correct option. [R is the gas constant.]

List-I

List-II

In the given P-V diagram, a monoatomic gas $(γ = \frac{5}{3})$ is first compressed adiabatically from state A to state B. Then it expands isothermally from state B to state C. [Given: $(\frac{1}{3})^{0.6} ≃ 0.5, ln 2 ≃ 0.7$ ]. Which of the following statement(s) is(are) correct?

The $p - V$ diagram of a diatomic ideal gas system going under cyclic process as shown in figure. The work done during an adiabatic process $C D$ is (use, $γ = 1.4$ )

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } State Coordinates

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Process AB: Differentials

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Process AB: Work Differential

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Process AB: Work Done

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Process AB: Heat Exchanged

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Process BC: Isobaric Heat

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Process BC: Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Process CA: Isothermal Heat

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Process CA: Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Cycle Summary