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 \(4\) cm\(^2\). The gas is heated slowly to raise the temperature by \(1.2\,^\circ\)C during which the piston moves by \(25\) mm. The amount of heat supplied to the gas is ________ J. (Atmospheric pressure \(=100\) kPa, \(R=8.3\) J/mol·K) (Neglect mass of the piston)

Using the first law of thermodynamics:
\(Q = \Delta U + W\)

Since the gas is diatomic with rotational modes only, the degrees of freedom are \(f = 5\), so:
\(C_V = \dfrac{5}{2}R\)

For \(n = 1\text{ mole}\) and \(\Delta T = 1.2\text{ K}\):
\(\Delta U = nC_V\Delta T = 1 \times \dfrac{5}{2} \times 8.3 \times 1.2 = 24.9\text{ J}\)

Work done by the gas:
\(W = P\,\Delta V\)

Cross-sectional area of the piston:
\(A = 4\text{ cm}^2 = 4 \times 10^{-4}\text{ m}^2\)

Piston displacement:
\(x = 25\text{ mm} = 2.5 \times 10^{-2}\text{ m}\)

Change in volume:
\(\Delta V = A \cdot x = (4 \times 10^{-4})(2.5 \times 10^{-2}) = 1 \times 10^{-5}\text{ m}^3\)

With \(P = 100\text{ kPa} = 10^5\text{ Pa}\):
\(W = (10^5)(1 \times 10^{-5}) = 1\text{ J}\)

Therefore, the heat supplied is:
\(Q = \Delta U + W = 24.9 + 1 = 25.9\text{ J}\)

Hence, the heat supplied to the gas is \(25.9\text{ J}\).