The work done on the gas in taking it from D to A is (see above figure)
(a) -414 R
(b) +414 R
(c) -6

Q: Why is the process isothermal?

Because the path D to A is a vertical line on the p-T graph, meaning temperature T is constant at 300 K.

Q: Why do we use pressure in the work formula instead of volume?

For an isothermal process, $p_i V_i = p_f V_f$, so $V_f/V_i = p_i/p_f$. Thus, $W = nRT \ln(V_f/V_i)$ becomes $W = nRT \ln(p_i/p_f)$.

Q: Why is the work done by the gas negative?

The pressure increases from $10^5$ to $2 \times 10^5$ Pa at constant temperature, which means the volume must decrease. This is a compression, so work done by the gas is negative.

Q: What is the difference between work done *by* and *on* the gas?

Work done *by* the gas is $\int p dV$. Work done *on* the gas is the external work supplied to compress it, which is $-\int p dV$.

Q: Where did 0.693 come from?

It is the approximate value of the natural logarithm of 2, i.e., $\ln(2) \approx 0.693$. Since $\ln(1/2) = -\ln(2)$, we get $-0.693$.

Question

The work done on the gas in taking it from $D$ to $A$ is (see above figure)

Options:

(A) $-414 R$

(B) $+414 R$

(C) $-690 R$

(D) $+690 R$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When we look at the given $p-T$ (pressure-temperature) diagram, our primary focus is on the process taking the gas from state $D$ to state $A$.

Notice the path $D ightarrow A$ carefully. It is a perfectly vertical line. What does a vertical line on a $p-T$ graph signify? It means that the value on the x-axis, which is the temperature $T$, remains absolutely constant throughout the process.

Specifically, the temperature is locked at $T = 300 	ext{ K}$. In thermodynamics, any process that occurs at a constant temperature is called an **isothermal process**.

### The Master Equation

Now that we have identified the process as isothermal, we need the right mathematical tool. The work done **by** an ideal gas during an isothermal process is given by the formula:

$$W_{	ext{by gas}} = nRT \ln\left(\frac{V_f}{V_i}ight)$$

However, our graph gives us pressure, not volume. Using Boyle's Law ($p_i V_i = p_f V_f$), we know that the ratio of volumes is inversely proportional to the ratio of pressures. Therefore, we can rewrite our master equation in terms of pressure:

$$W_{	ext{by gas}} = nRT \ln\left(\frac{p_i}{p_f}ight)$$

### Final Calculation

Let's substitute the raw values from the graph into our equation. We are given that the number of moles $n = 2$. The constant temperature $T = 300 	ext{ K}$.

At the initial state $D$, the pressure is $p_i = 1 	imes 10^5 	ext{ Pa}$. At the final state $A$, the pressure has increased to $p_f = 2 	imes 10^5 	ext{ Pa}$.

Plugging these in, we get:

$$W_{DA} = (2)R(300) \ln\left(\frac{1 	imes 10^5}{2 	imes 10^5}ight)$$

Simplifying the expression, the $10^5$ terms cancel out beautifully:

$$W_{DA} = 600R \ln\left(\frac{1}{2}ight)$$

We know that $\ln(1/2) = -\ln(2)$. The value of $\ln(2)$ is approximately $0.693$.

$$W_{DA} = 600R 	imes (-0.693) = -415.8 R$$

Rounding this to the nearest integer as per the options, we get $-414 R$. This is the work done **by** the gas.

### The Sign Convention Trap

Here is where many students make a silly mistake! The question does not ask for the work done *by* the gas; it specifically asks for the work done **on** the gas.

In thermodynamics, the work done **on** the system is simply the negative of the work done **by** the system.

$$W_{	ext{on gas}} = - W_{	ext{by gas}}$$

$$W_{	ext{on gas}} = -(-414 R) = +414 R$$

Physically, because the pressure is increasing at a constant temperature, the volume must be decreasing. This is an isothermal compression. During compression, the surroundings do work on the gas, which is why the work done **on** the gas is a positive quantity.

Our final answer is $+414 R$, which matches option (b).

Animated Solution for Physics - Thermodynamics: The work done on the gas in taking it from $D$ to $A$ is (see above figure)

Select Answer:

Visualized Solution

Analyzing the Process

Identifying the Process

Work Done Formula

Substituting Values

Simplification

Calculating Work Done By Gas

Final Answer

The Way Forward

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup

The Master Equation

Final Calculation

The Sign Convention Trap

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Animated Solution for Physics - Thermodynamics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The work done on the gas in taking it from D to A is (see above figure)