Correct Answer: 15.5

Explanation

The gas is heated slowly with a frictionless piston, hence the process is isobaric.

For a monoatomic gas, change in internal energy is given by

$$ \Delta U = 3nR\Delta T $$

Here $n = 1$ and $\Delta T = 4\ \text{K}$.

$$ \Delta U = 3R \times 4 = 12R $$

Heat supplied is used in increasing internal energy and doing work:

$$ Q = \Delta U + W $$

$$ 126 = 12R + W $$

Using $R = 8.3\ \text{J mol}^{-1}\text{K}^{-1}$,

$$ \Delta U = 12 \times 8.3 = 99.6\ \text{J} $$

$$ W = 126 - 99.6 = 26.4\ \text{J} $$

Work done at constant pressure:

$$ W = P\Delta V $$

$$ \Delta V = \frac{26.4}{10^5} = 2.64 \times 10^{-4}\ \text{m}^3 $$

Area of piston:

$$ A = 17\ \text{cm}^2 = 17 \times 10^{-4}\ \text{m}^2 $$

Displacement of piston:

$$ x = \frac{\Delta V}{A} = \frac{2.64 \times 10^{-4}}{17 \times 10^{-4}} \approx 0.155\ \text{m} $$

$$ x \approx 15.5\ \text{cm} $$

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