One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in the figure.

Q: Why is process AB isothermal?

In the given $V-T$ graph, the line representing process $AB$ is vertical. This means the temperature $T$ remains constant at $T_0$ while the volume changes. A process with constant temperature is called an isothermal process.

Q: Why are the internal energies at A and B the same?

The internal energy of an ideal gas is directly proportional to its absolute temperature ($U = \frac{f}{2}nRT$). Since process $AB$ is isothermal, $T_A = T_B$, which implies $U_A = U_B$.

Q: How did we calculate the work done in process AB?

For an isothermal process, the work done is given by $W = nRT \ln(V_f/V_i)$. Substituting $n=1$, $T=T_0$, $V_i=V_0$, and $V_f=4V_0$, we get $W = RT_0 \ln 4$. Using the ideal gas law at state A ($p_0V_0 = RT_0$), this becomes $p_0V_0 \ln 4$.

Q: Why can't we assume process BC is isobaric?

An isobaric process on a $V-T$ graph is represented by a straight line that passes through the origin ($V \propto T$). Although $BC$ is a straight line, the graph does not provide enough information to confirm that its extension passes through the origin.

Q: What if line BC did pass through the origin?

If $BC$ passed through the origin, it would be an isobaric process. We could then use $p_B = p_C$. From the isothermal process $AB$, $p_B = p_0/4$. Thus, $p_C$ would be $p_0/4$, and using $V_C/T_C = V_B/T_B$, $T_C$ would be $T_0/4$, making options (c) and (d) correct.

Question

One mole of an ideal gas in initial state $A$ undergoes a cyclic process $ABCA$, as shown in the figure. Its pressure at $A$ is $p_0$. Choose the correct option(s) from the following.

Options:

(A) Internal energies at $A$ and $B$ are the same

(B) Work done by the gas in process $AB$ is $p_0 V_0 \ln 4$

(C) Pressure at $C$ is $\frac{p_0}{4}$

(D) Temperature at $C$ is $\frac{T_0}{4}$

Sigma Percentile · Accepted Answer

Welcome to this fascinating journey through thermodynamics! Today, we are going to dissect a beautiful problem involving a cyclic process of an ideal gas. At first glance, graphs can be deceiving, but by carefully analyzing the axes and the physical laws governing the gas, we can unravel the entire story. Let's dive in!

### Decoding the V-T Graph

The very first step in tackling any graphical thermodynamics problem is to **look at the axes**. It is a common pitfall to assume every graph is a $p-V$ (pressure-volume) diagram. Here, the vertical axis represents Volume ($V$) and the horizontal axis represents Temperature ($T$).

We are given one mole of an ideal gas undergoing a cyclic process $ABCA$. Let's trace its path. The process starts at state $A$, where the temperature is $T_0$ and the volume is $V_0$. The pressure at this initial state is given as $p_0$.

### The Isothermal Journey from A to B

Look closely at the line connecting state $A$ to state $B$. It is a perfectly vertical line. On a $V-T$ graph, a vertical line means that the temperature is not changing. Therefore, $T_A = T_B = T_0$. A process where the temperature remains constant is called an **isothermal process**.

Now, what does this mean for the internal energy of the gas? The internal energy ($U$) of an ideal gas is a function of its absolute temperature alone, given by the relation $U = \frac{f}{2}nRT$. Since the temperature at $A$ and $B$ is exactly the same, their internal energies must also be identical. This confirms that **Internal energies at $A$ and $B$ are the same**.

Next, let's calculate the work done during this isothermal expansion. The gas expands from an initial volume $V_0$ to a final volume $4V_0$. The work done in an isothermal process is given by the elegant formula:

$$W = nRT \ln\left(\frac{V_f}{V_i}ight)$$

Substituting our known values ($n=1$, $T=T_0$, $V_i=V_0$, $V_f=4V_0$), we get:

$$W_{AB} = RT_0 \ln\left(\frac{4V_0}{V_0}ight) = RT_0 \ln 4$$

We can express this in terms of the initial pressure and volume. Using the ideal gas equation at state $A$, we know that $p_A V_A = nRT_A$, which translates to $p_0 V_0 = RT_0$. Substituting this back into our work equation, we find:

$$W_{AB} = p_0 V_0 \ln 4$$

This perfectly matches our second option!

### The Ambiguity of Process BC

Now, let's move from state $B$ to state $C$. The graph shows a straight line connecting these two points. Here is where many students fall into a trap. It is tempting to assume that this straight line represents an isobaric (constant pressure) process.

For a process to be isobaric on a $V-T$ graph, the volume must be directly proportional to the temperature ($V \propto T$), which means the straight line **must pass through the origin**.

If we assume the line $BC$ passes through the origin, we could calculate the pressure at $C$. Since $AB$ is isothermal, $p_A V_A = p_B V_B$, giving us $p_B = \frac{p_0}{4}$. An isobaric process would mean $p_C = p_B = \frac{p_0}{4}$. Furthermore, using Charles's Law ($V/T = 	ext{constant}$), we would find $T_C = \frac{T_0}{4}$.

However, **the graph does not explicitly show or state that the line $BC$ passes through the origin**. In physics, we cannot make assumptions based on visual approximations unless explicitly stated. Because of this crucial missing piece of information, we cannot definitively determine the pressure and temperature at state $C$.

Thus, we must conclude that only the first two statements are verifiably correct. This problem is a brilliant reminder to always read graphs carefully and never assume information that isn't explicitly provided!

Animated Solution for Physics - Thermodynamics: 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.

Select Answer:

Visualized Solution

$T_{A} = T_{B}$

$W_{A B} = n R T ln (\frac{V _{f}}{V _{i}})$

$W_{A B} = p_{0} V_{0} ln 4$

$V \propto T$

$p_{C} \neq = \frac{p _{0}}{4}$

The Sigma Insight: Thermodynamic Processes

Decoding the V-T Graph

The Isothermal Journey from A to B

The Ambiguity of Process BC

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

One mole of a monoatomic ideal gas undergoes a cyclic process as shown in the figure (where, $V$ is the volume and $T$ is the temperature). Which of the statements below is (are) true ?

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]