The work performed by a sample of an ideal gas during process A is twice the work done during another process B . The temperature increases by an identical amount in both processes. If C_A and C_B denote the molar heat capacities for these two processes, then

Step-by-step solution

According to the first law of thermodynamics: Q = U + W For an ideal gas, the change in internal energy depends only on the change in temperature, so U = n C_V T . The molar heat capacity for any process is given by C = Q n T . Substituting the expression for Q , we get: C = U + W n T = C_V + W n T For process A and process B , the molar heat capacities are: C_A = C_V + W_A n T C_B = C_V + W_B n T Subtracting the two equations yields: C_A - C_B = W_A - W_B n T It is given that the work performed by the gas in proce