The internal energy of a monoatomic gas is 3nRT. One mole of helium is kept in a cylinder having internal cross section area of 17 $$cm^{2}$$ and fitted with a light movable frictionless piston. The gas is heated slowly by suppling 126 J heat. If the temperature rises by $$4^{o}C$$, then the piston will move ____ cm.
(atmospheric pressure= $$10^{5}$$ Pa)

The first law of thermodynamics states
$$Q = \Delta U + W$$
where $$Q$$ is the heat supplied, $$\Delta U$$ is the change in internal energy and $$W$$ is the work done by the gas.

The question itself specifies that for a mono-atomic gas
$$U = 3\,nRT$$
Hence the change in internal energy is
$$\Delta U = 3\,nR\Delta T$$

Given data:
• Number of moles, $$n = 1$$
• Universal gas constant, $$R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$$
• Rise in temperature, $$\Delta T = 4\,\text{K}$$
• Heat supplied, $$Q = 126\ \text{J}$$

Calculate $$\Delta U$$:
$$\Delta U = 3 \times 1 \times 8.314 \times 4 = 99.768\ \text{J}$$

Find the work done using the first law:
$$W = Q - \Delta U = 126 - 99.768 = 26.232\ \text{J}$$

The cylinder is fitted with a light, frictionless piston, so the gas expands slowly against the constant atmospheric pressure
$$P = 10^{5}\ \text{Pa}$$
For a constant external pressure
$$W = P\,\Delta V \;\; \Longrightarrow \;\; \Delta V = \frac{W}{P}$$
$$\Delta V = \frac{26.232}{10^{5}} = 2.6232 \times 10^{-4}\ \text{m}^{3}$$

The piston’s cross-sectional area is
$$A = 17\ \text{cm}^{2} = 17 \times 10^{-4}\ \text{m}^{2} = 1.7 \times 10^{-3}\ \text{m}^{2}$$

Let the piston move up by a distance $$x$$. The change in volume is also
$$\Delta V = A\,x$$
Therefore
$$x = \frac{\Delta V}{A} = \frac{2.6232 \times 10^{-4}}{1.7 \times 10^{-3}} = 0.1543\ \text{m}$$

Converting to centimetres:
$$x = 0.1543\ \text{m} \times 100 = 15.43\ \text{cm} \approx 15.5\ \text{cm}$$

Hence the piston will move about 15.5 cm.
Option D is correct.