40 g of a gas is heated at constant pressure, resulting in a temperature rise of 50^ C . The specific heat of the gas at constant pressure is 0.25 kcal kg ⁻¹ ^ C ⁻¹ and at constant volume is 0.17 kcal kg ⁻¹ ^ C ⁻¹ . Given 1 cal = 4.2 J , the work done by the gas during this expansion is:

Step-by-step solution

According to the First Law of Thermodynamics, the heat supplied at constant pressure is: Q = U + W The work done by the gas is: W = Q - U We know that Q = m c_p T and U = m c_v T . Therefore: W = m (c_p - c_v) T Given: m = 40 g = 0.04 kg T = 50^ C c_p = 0.25 kcal kg ⁻¹ ^ C ⁻¹ c_v = 0.17 kcal kg ⁻¹ ^ C ⁻¹ Difference in specific heats: c_p - c_v = (0.25 - 0.17) kcal kg ⁻¹ ^ C ⁻¹ = 0.08 kcal kg ⁻¹ ^ C ⁻¹ Converting to J kg ⁻¹ ^ C ⁻¹ : c_p - c_v = 0.08 1000 4.2 J kg ⁻¹ ^ C ⁻¹ = 336 J kg ⁻¹ ^ C ⁻¹ Substituting the value