Exams › JEE Advanced › Chemistry

One mole of a monatomic ideal gas at initial pressure 2.00 bar and temperature 273 K undergoes a reversible process defined by P/V = constant, ending at a final pressure of 4.00 bar. Given Cv,m = (3/2)R, find the ratio delta_U / W (where W is the work done by the gas).

1. -3.0
2. -1.5
3. +1.5
4. +3.0

Correct answer: +3.0

Solution

Since P/V = k (constant), P = kV. Using ideal gas law PV = RT (n=1 mole): P(P/k) = RT => P² = kRT, so T is proportional to P². T2/T1 = (P2/P1)² = (4/2)² = 4, giving T2 = 4 x 273 = 1092 K. delta_U = Cv,m x delta_T = (3/2)R x (1092-273) = (3/2)R x 819. Polytropic index n_poly = -1: W = R(T1 - T2)/(n_poly - 1) = R(273 - 1092)/(-1 - 1) = R(-819)/(-2) = 819R/2. Ratio delta_U/W = (3/2)x819R / (819R/2) = (3/2)/(1/2) = 3. So delta_U/W = +3.0.

Related JEE Advanced Chemistry questions

• If the specific heat at constant pressure (Cp) is 10 cal at 1000 K, the initial temperature (T1) is 1000 K, the final temperature (T2) is 100 K, and the mass (m) is 32 g, what is the entropy change (ΔS) under constant pressure?
• For the reaction H2(g) + 1/2 O2(g) → H2O(l), where the enthalpy of formation (ΔHf) is given as −285.9 kJ mol⁻¹, determine the value of ΔHf.
• For the reaction C3H8(g) → 3C(g) + 8H(g), determine the enthalpy change (ΔHf) given that ΔHf for C2H2(g) → 2C(g) + 6H(g) is −620 kJ mol⁻¹.
• The standard enthalpies of formation of CO₂(g), H₂O(l) and glucose(s) at 25°C are -400 kJ/mol, -300 kJ/mol and -1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25°C is -
• Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is(are) -
• During the transformation of liquid water to water vapor at 100°C and standard atmospheric pressure, which of the following is accurate?

⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →