Thermodynamics Work from PV Graph

 

❓ Question

A PV curve follows the equation

$$4aP=(V-2)^2$$

Find the net work done by the gas from

$$V=1 \quad \text{to} \quad V=3$$

where $a$ is a constant.

Options:

1. $\frac{a}{3}$
2. $\frac{a}{6}$
3. $-\frac{a}{3}$
4. $-\frac{a}{6}$

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✍️ Short Explanation

Work done by a gas:

$$W=\int P\,dV$$

Given:

$$4aP=(V-2)^2$$
$$P=\frac{(V-2)^2}{4a}$$

So,

$$W=\frac1{4a}\int_{1}^{3}(V-2)^2\,dV$$

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🔷 Step 1 — Substitute

Let

$$u=V-2$$

Then:

• $V=1 \Rightarrow u=-1$
  
• $V=3 \Rightarrow u=1$
  

Hence

$$W=\frac1{4a}\int_{-1}^{1}u^2\,du$$

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🔷 Step 2 — Integrate

$$\int_{-1}^{1}u^2\,du = \left[\frac{u^3}{3}\right]_{-1}^{1} = \frac13-\left(-\frac13\right) = \frac23$$

Therefore,

$$W=\frac1{4a}\cdot\frac23 = \frac1{6a}$$

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✅ Final Answer

$$\boxed{\frac{1}{6a}}$$

Option (2)

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🔷 JEE Shortcut

Since $(V-2)^2$ is symmetric about $V=2$,

$$W=\frac1{4a}\int_{-1}^{1}u^2\,du$$

and

$$\int_{-1}^{1}u^2\,du=\frac23$$

giving immediately:

$$\boxed{\frac1{6a}}$$

in a few seconds.

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