The internal energy of one mole of an ideal gas at temperature T is given by U = 3RT/2. The process AB is isothermal i.e., T_A = T_B = T₀, which makes U_A = U_B.

The work done in the isothermal process AB is

W_AB = nRT ln(V_B/V_A)

= nRT₀ ln(4V₀/V₀) = p₀V₀ ln4.

Information regarding p and T at C can not be obtained from the given graph. Unless it is mentioned that line BC passes through origin or not.

Hence, the correct options are (a) and (b).

Note:

If we assume that the line BC pass through the origin. In the process BC, the slope V/T is constant. Thus, BC is an isobaric process (as V/T = nR/p, by the ideal gas equation). Thus,

p_C = p_B = RT_B / V_B

= RT₀/(4V₀) = (RT₀/V₀)4 = p₀/4.

Apply the ideal gas equation for the states A and C to get

$${T_C} = {{{p_C}{V_C}} \over {{p_0}{V_0}}}{T_0} = {{({p_0}/4){V_0}} \over {{p_0}{V_0}}}{T_0} = {T_0}/4$$.