The internal energy of a monatomic ideal gas is 1.5 nRT. One mole of helium is kept in a cylinder of cross-section 8.5 ${\mathrm{cm}}^2$. The cylinder is closed by a light frictionless piston. The gas is heated slowly in a process during which a total of 42 J heat is given to the gas. If the temperature rises through $2^{\mathrm{o}}\mathrm{C}$, find the distance moved by the piston. Atmospheric pressure = 100 kPa.

The change in internal energy of the gas is $∆\mathrm{U} = 1.5 \mathrm{nR} (∆\mathrm{T})$

                                                                              = $1.5(1 \mathrm{mol})(8.3 \mathrm{J}/\mathrm{mol}-\mathrm{K})(2 \mathrm{K}) = 24.9 \mathrm{J}$

The heat given to the gas = 42 J

The work done by the gas is $∆\mathrm{W} = ∆\mathrm{Q} - ∆\mathrm{U} = 42\mathrm{J} - 24.9 \mathrm{J} = 17.1 \mathrm{J}$

If the distance moved by the piston is x, the work done is $∆\mathrm{W} = (100 \mathrm{kPa})(8.5 {\mathrm{cm}}^2)\mathrm{x}$

Thus, $({10}^5 \mathrm{N}/{\mathrm{m}}^2)(8.5 \times {10}^{-4}{\mathrm{m}}^2)\mathrm{x} = 17.1\mathrm{J}$

or x = 0.2 m = 20 cm