A monoatomic ideal gas of two moles is taken through a cyclic process starting from A as shown in th

Q: Why is process A to B isobaric?

On a V-T graph, a straight line passing through the origin means $V \propto T$. According to the ideal gas law ($pV = nRT$), this is only possible if the pressure $p$ is constant.

Q: Why do we use $C_p$ for process A to B and $C_V$ for process C to D?

Process A to B occurs at constant pressure, so the heat exchanged is calculated using the molar heat capacity at constant pressure ($C_p$). Process C to D occurs at constant volume, so we use the molar heat capacity at constant volume ($C_V$).

Q: How did you know the temperature at D is $300\text{ K}$?

Process D to A is a vertical line on the V-T graph, which means the temperature is constant. Since the temperature at A is given as $300\text{ K}$, the temperature at D must also be $300\text{ K}$.

Q: Why is the change in internal energy zero for the complete cycle?

Internal energy is a state function, meaning it depends only on the current state of the system, not the path taken. In a complete cycle, the system returns to its initial state, so the net change in internal energy is exactly zero.

Q: Why did the $\ln(2)$ terms cancel out in the final calculation?

The heat absorbed during the isothermal expansion (B to C) depends on $\ln(V_C/V_B) = \ln(2)$. The heat released during the isothermal compression (D to A) depends on $\ln(V_A/V_D) = \ln(1/4) = -2\ln(2)$. Since the temperature at B is twice the temperature at A, the magnitudes of these heat exchanges perfectly balance out the log terms!

Question

A monoatomic ideal gas of two moles is taken through a cyclic process starting from $A$ as shown in the figure. The volume ratio are $\frac{V_B}{V_A} = 2$ and $\frac{V_D}{V_A} = 4$. If the temperature $T_A$ at $A$ is $27^\circ	ext{C}$.
Calculate
(a) the temperature of the gas at point $B$,
(b) heat absorbed or released by the gas in each process,
(c) the total work done by the gas during the complete cycle.
Express your answer in terms of the gas constant $R$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome to a fascinating journey through a thermodynamic cycle! We are given a $V-T$ (Volume-Temperature) graph for $2$ moles of a monoatomic ideal gas. The cycle consists of four distinct paths: $A ightarrow B$, $B ightarrow C$, $C ightarrow D$, and $D ightarrow A$.

Before we dive into the calculations, let's establish our baseline parameters. For a monoatomic gas, the molar heat capacities are $C_V = \frac{3}{2}R$ and $C_p = \frac{5}{2}R$. We are given the initial temperature $T_A = 27^\circ	ext{C} = 300	ext{ K}$. The volume ratios are $V_B = 2V_A$ and $V_D = V_C = 4V_A$.

### Process A to B: The Isobaric Expansion

Look closely at the path from $A$ to $B$. It is a straight line passing through the origin. Geometrically, this implies that the volume is directly proportional to the temperature ($V \propto T$). According to the ideal gas law ($pV = nRT$), this proportionality holds true only if the pressure $p$ remains constant. Thus, $A ightarrow B$ is an **isobaric process**.

Since $V \propto T$, the ratio of volume to temperature is constant. We can easily find the temperature at $B$:

$$\frac{T_B}{T_A} = \frac{V_B}{V_A} = 2$$

$$\Rightarrow T_B = 2T_A = 2 	imes 300	ext{ K} = 600	ext{ K}$$

Now, let's calculate the heat exchanged during this isobaric expansion. The formula for heat at constant pressure is $Q_{AB} = nC_p \Delta T$:

$$Q_{AB} = (2)\left(\frac{5}{2}Right)(T_B - T_A)$$

$$Q_{AB} = 5R(600 - 300) = 1500 R$$

The positive sign indicates that $1500 R$ of heat is **absorbed** by the gas.

### Process B to C: The Isothermal Expansion

Next, observe the path from $B$ to $C$. It is a vertical line on the $V-T$ graph, which means the temperature remains constant at $T_B = 600	ext{ K}$. This is an **isothermal process**.

In an isothermal process, the change in internal energy is zero ($\Delta U = 0$), so the heat absorbed equals the work done by the gas. The formula is $Q_{BC} = W_{BC} = nRT_B \ln\left(\frac{V_C}{V_B}ight)$:

$$Q_{BC} = (2)R(600) \ln\left(\frac{4V_A}{2V_A}ight)$$

$$Q_{BC} = 1200 R \ln(2) \approx 831.6 R$$

Again, the positive sign means heat is **absorbed**.

### Process C to D: The Isochoric Cooling

Moving on to the path from $C$ to $D$, we see a horizontal line. This indicates that the volume is constant at $V_D = 4V_A$. This is an **isochoric process**.

The heat exchanged at constant volume is given by $Q_{CD} = nC_V \Delta T$. The temperature drops from $T_B$ back to $T_A$:

$$Q_{CD} = (2)\left(\frac{3}{2}Right)(T_D - T_C)$$

$$Q_{CD} = 3R(300 - 600) = -900 R$$

The negative sign tells us that $900 R$ of heat is **released** by the gas.

### Process D to A: The Isothermal Compression

Finally, the path from $D$ back to $A$ is another vertical line, meaning it is an **isothermal process** at a constant temperature of $T_A = 300	ext{ K}$.

Just like before, the heat exchanged equals the work done:

$$Q_{DA} = W_{DA} = nRT_A \ln\left(\frac{V_A}{V_D}ight)$$

$$Q_{DA} = (2)R(300) \ln\left(\frac{V_A}{4V_A}ight)$$

$$Q_{DA} = 600 R \ln\left(\frac{1}{4}ight) = -1200 R \ln(2) \approx -831.6 R$$

Heat is **released** during this compression phase.

### The Grand Finale: Net Work Done

We have successfully calculated the heat exchanged in all four processes. For any complete thermodynamic cycle, the net change in internal energy is zero ($\Delta U_{	ext{cycle}} = 0$). According to the first law of thermodynamics, this means the total net work done by the gas equals the net heat exchanged:

$$W_{	ext{net}} = Q_{	ext{net}} = Q_{AB} + Q_{BC} + Q_{CD} + Q_{DA}$$

$$W_{	ext{net}} = 1500 R + 1200 R \ln(2) - 900 R - 1200 R \ln(2)$$

Notice how beautifully the logarithmic terms perfectly cancel each other out! We are left with a clean, elegant integer result:

$$W_{	ext{net}} = 1500 R - 900 R = 600 R$$

And there you have it! By carefully decoding the geometry of the $V-T$ graph, we've unraveled the entire thermodynamic cycle.

Visualized Solution

$V - T Graph Analysis$

$Process A \to B$

$Temperature at B$

$Calculating T_{B}$

$Heat in Process A \to B$

$Calculating Q_{A B}$

$Process B \to C$

$Heat in Process B \to C$

$Process C \to D$

$Heat in Process C \to D$

$Process D \to A$

$Heat in Process D \to A$

$Complete Cycle$

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup

Process A to B

Process B to C

Process C to D

Process D to A

The Grand Finale

Similar Questions

One mole of a monoatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature (V-T) diagram. The correct statement(s) is/are: [R is the gas constant]

One mole of an ideal gas in initial state $A$ undergoes a cyclic process $A B C A$ , as shown in the figure. Its pressure at $A$ is $p_{0}$ . Choose the correct option(s) from the following.

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

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V−T Graph Analysis

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat in Process A→B

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat in Process B→C

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat in Process C→D

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat in Process D→A

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