Direction (Q.Nos. 49 to 51) are based on the following figure. Assume the gas to be ideal, the work

Q: Why did we assume the process is isobaric?

Look at the $p-T$ graph. The line connecting state A to state B is perfectly horizontal. This means the value on the y-axis (pressure) remains constant at $2 \times 10^5 \text{ Pa}$ throughout the process. A constant pressure process is called isobaric.

Q: The formula for work is $p\Delta V$. Why did we use $nR\Delta T$?

You are absolutely right! However, the graph doesn't give us the volume. For an ideal gas, $pV = nRT$. When pressure is constant, any change in volume is directly proportional to the change in temperature, so $p\Delta V = nR\Delta T$. This clever substitution lets us solve the problem using the given temperatures.

Q: The question asks for work done *on* the gas. Shouldn't the answer be negative?

Technically, yes. Since the gas expands (temperature increases at constant pressure), it does positive work *by* the gas. The work done *on* the gas would be $-400 R$. However, since all the given options are positive, the question is asking for the magnitude of the work. Always adapt to the options provided!

Q: What is the significance of $\gamma = 1.67$ given in the problem?

For this specific part of the question (calculating work from A to B), $\gamma$ is not needed. It is extra information that might be required for other parts of the linked comprehension question, such as calculating the change in internal energy ($\Delta U = nC_V\Delta T$) or heat supplied.

Q: What if the process was from A to D instead?

The line from A to D is vertical, meaning the temperature is constant at $300 \text{ K}$. This would be an isothermal process. The work done in an isothermal process is calculated differently, using the formula $W = nRT \ln(p_1/p_2)$.

Question

Direction (Q.Nos. 49 to 51) are based on the following figure.
Assume the gas to be ideal, the work done on the gas in taking it from $A$ to $B$ is

Options:

(A) $200 R$

(B) $300 R$

(C) $400 R$

(D) $500 R$

Sigma Percentile · Accepted Answer

### Decoding the Pressure-Temperature Graph

Imagine you are an engineer analyzing the performance of a heat engine. The first step is always to understand the thermodynamic cycle it undergoes. In this problem, we are presented with a pressure-temperature ($p-T$) graph, which is a bit different from the usual pressure-volume ($p-V$) graphs you might be used to.

Our focus is specifically on the process taking the gas from state $A$ to state $B$. Let's look closely at the line connecting these two states. It is a perfectly horizontal line.

What does a horizontal line on a $p-T$ graph tell us? It means that the value on the y-axis, which is pressure, does not change. Therefore, the pressure remains constant at $2 	imes 10^5 	ext{ Pa}$ throughout the process.

**A process that occurs at a constant pressure is known as an isobaric process.**

Simultaneously, as we move from $A$ to $B$, we can read the corresponding temperatures on the x-axis. The temperature increases from an initial value of $T_A = 300 	ext{ K}$ to a final value of $T_B = 500 	ext{ K}$.

### The Master Equation for Work Done

Now that we have identified the process as isobaric, we need to calculate the work done. The fundamental definition of work done by a gas at constant pressure is given by:

$$W = p\Delta V$$

Here, $\Delta V$ represents the change in volume. But wait, our graph doesn't provide any information about the volume! How can we proceed?

This is where the **Ideal Gas Equation** comes to our rescue as a powerful bridge. For an ideal gas, we know that:

$$pV = nRT$$

Since the pressure $p$ is constant, any change in volume $\Delta V$ must be directly proportional to a change in temperature $\Delta T$. Mathematically, we can write this relationship as:

$$p\Delta V = nR\Delta T$$

This is a brilliant substitution! It allows us to calculate the work done using temperature, which we *do* know, instead of volume, which we *don't*. Our new working formula becomes:

$$W = nR\Delta T = nR(T_B - T_A)$$

### Executing the Final Calculation

We are now fully equipped to find the numerical answer. Let's gather all the known values from the problem statement and the graph.

We are given that the number of moles of the gas is $n = 2$. The initial temperature is $T_A = 300 	ext{ K}$, and the final temperature is $T_B = 500 	ext{ K}$.

Let's substitute these values into our master equation:

$$W = 2 	imes R 	imes (500 - 300)$$

First, we calculate the temperature difference:

$$\Delta T = 500 - 300 = 200 	ext{ K}$$

Now, multiply this by the number of moles:

$$W = 2 	imes R 	imes 200$$

$$W = 400 R$$

### A Note on Sign Conventions

We have calculated the work done *by* the gas to be $400 R$. Because the temperature increased at constant pressure, the volume must have also increased (expansion). When a gas expands, it does positive work on its surroundings.

The question specifically asks for the work done **on** the gas. Strictly speaking, the work done *on* the gas is the negative of the work done *by* the gas, which would be $-400 R$.

However, in many multiple-choice questions, if all the options are positive, the examiners are simply asking for the **magnitude** of the work involved. Looking at our options, $400 R$ perfectly matches option (c).

By carefully analyzing the graph and cleverly using the ideal gas law, we've arrived at the correct solution with confidence!

Animated Solution for Physics - Thermodynamics: Direction (Q.Nos. 49 to 51) are based on the following figure. Assume the gas to be ideal, the work done on the gas in taking it from $A$ to $B$ is

Select Answer:

Visualized Solution

Analyzing Process $A \to B$

Work Done in Isobaric Process

Substituting the Values

Calculating the Work Done

Final Answer

The Sigma Insight: Thermodynamic Processes

Decoding the Pressure-Temperature Graph

The Master Equation for Work Done

Executing the Final Calculation

A Note on Sign Conventions

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