One mole of diatomic gas having rotational modes only is kept in a cylinder with a piston system. The cross-section area of the cylinder is $4\ \text{cm}^2$. The gas is heated slowly to raise the temperature by $1.2^\circ\text{C}$ during which the piston moves by $25$ mm. The amount of heat supplied to the gas is ______ J.
(Atmospheric pressure $= 100\ \text{kPa}$, $R = 8.3\ \text{J/mol.K}$) (Neglect mass of the piston)
Answer: (B) $10.96$
The piston is massless, so the gas pushes against atmospheric pressure. Work done by the gas:
$$W = p\,A\,\Delta x = 10^5 \times 4 \times 10^{-4} \times 0.025 = 1.0\ \text{J}$$
The official answer counts only the two rotational modes ("rotational modes only") in the internal energy: $\Delta U = n\dfrac{f}{2}R\Delta T$ with $f = 2$,
$$\Delta U = 1 \times 8.3 \times 1.2 = 9.96\ \text{J}$$
First law: $Q = \Delta U + W = 9.96 + 1.0 = 10.96$ J.
Note: with the usual $5$ degrees of freedom of a diatomic gas, $\Delta U = 24.9$ J, which matches no option. The data are also not fully consistent, since for an ideal gas at constant pressure $p\Delta V$ should equal $nR\Delta T = 9.96$ J.
Solution by Sreeraj P, M.Sc Physics