One mole of an ideal gas at 27°C is taken from A to B as shown in the given p-V indicator diagram. T

Q: How did we know it's an isothermal process just by looking at the graph?

By checking the coordinates! At point A, $p \times V = 200 \times 2 = 400$. At point B, $p \times V = 100 \times 4 = 400$. Since $pV$ is constant, the temperature must be constant according to the ideal gas law.

Q: Why do we use temperature in Kelvin instead of Celsius?

In all thermodynamic equations derived from the ideal gas law ($pV = nRT$), the temperature $T$ represents the absolute temperature, which must strictly be in Kelvin.

Q: What if the curve was a straight line connecting A and B?

If it were a straight line, the process would not be isothermal. The work done would simply be the area of the trapezium under the line, calculated as $\frac{1}{2} \times (p_A + p_B) \times (V_B - V_A)$.

Q: Why is the work done positive?

The volume of the gas is increasing from $2 \text{ m}^3$ to $4 \text{ m}^3$. Whenever a gas expands, it does positive work on its surroundings.

Q: How do I handle the $10^{-1}$ format at the end?

You need to shift the decimal point of your final answer one place to the right to compensate for the $10^{-1}$ multiplier. So, $1725.8$ becomes $17258 \times 10^{-1}$.

Question

One mole of an ideal gas at $27^{\circ}	ext{C}$ is taken from $A$ to $B$ as shown in the given $p-V$ indicator diagram. The work done by the system will be ...... $	imes 10^{-1}	ext{ J}$.
[Take, $R = 8.3	ext{ J/mol-K}$, $\ln 2 = 0.6931$]
(Round off to the nearest integer)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem presents us with a $p-V$ indicator diagram where an ideal gas is taken from state $A$ to state $B$. The first crucial step in any thermodynamics problem involving a $p-V$ graph is to identify the nature of the process.

Let's look at the coordinates of the initial and final states. At point $A$, the pressure is $p_A = 200 	ext{ N/m}^2$ and the volume is $V_A = 2 	ext{ m}^3$. The product of pressure and volume here is $p_A V_A = 400 	ext{ J}$.

At point $B$, the pressure drops to $p_B = 100 	ext{ N/m}^2$ while the volume increases to $V_B = 4 	ext{ m}^3$. Calculating the product again, we get $p_B V_B = 400 	ext{ J}$.

Since $p_A V_A = p_B V_B$, the product $pV$ is constant throughout the process. According to the ideal gas law, $pV = nRT$, a constant $pV$ implies that the temperature $T$ remains constant. Therefore, the curve from $A$ to $B$ represents an **isothermal expansion**.

### The Master Equation

Now that we have identified the process as isothermal, we need to calculate the work done by the gas. Geometrically, the work done is the area under the $p-V$ curve. For an isothermal process, the work done is given by the standard formula:

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

This is our master equation. It elegantly connects the macroscopic work done to the initial and final volumes, along with the constant temperature of the system.

### Substituting and Calculating

Let's gather all the given values to plug into our equation. We are given $n = 1 	ext{ mole}$ of an ideal gas. The universal gas constant is $R = 8.3 	ext{ J/mol-K}$. The temperature is given as $27^\circ	ext{C}$, which we must strictly convert to Kelvin for thermodynamic calculations: $T = 27 + 273 = 300 	ext{ K}$. The volume expands from $V_i = 2 	ext{ m}^3$ to $V_f = 4 	ext{ m}^3$.

Substituting these into our master equation:

$$W = 1 	imes 8.3 	imes 300 	imes \ln\left(\frac{4}{2}ight)$$

$$W = 2490 	imes \ln(2)$$

The problem provides the value of $\ln(2) = 0.6931$.

$$W = 2490 	imes 0.6931 = 1725.819 	ext{ J}$$

This is the exact work done by the gas during the expansion.

### The Final Formatting Trap

We have the work done, but we must be extremely careful with how the question asks for the final answer. It requires the answer in the format of $... 	imes 10^{-1} 	ext{ J}$.

To match this format, we rewrite our result:

$$1725.819 	ext{ J} = 17258.19 	imes 10^{-1} 	ext{ J}$$

The question also instructs us to round off to the nearest integer. Rounding $17258.19$ gives us **$17258$**.

This is a classic JEE trap where students might do all the physics correctly but lose marks due to a formatting oversight. Always read the final line of the question twice!

Enter Numerical Value:

Visualized Solution

$Analyzing the p - V Diagram$

$Work Done in Isothermal Process$

$Substituting the Values$

$Simplifying the Expression$

$Calculating Final Value$

$Formatting the Answer$

$What if the path was a straight line?$

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup

The Master Equation

Substituting and Calculating

The Final Formatting Trap

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the p−V Diagram

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if the path was a straight line?