Thermodynamics practice questions.

2.0 mol of an ideal monatomic gas is heated at constant volume from 300 K to 350 K. Determine the change in internal energy of the gas.

• 0.83 kJ
• 8.7 kJ
• 1.2 kJ
• 2.1 kJ

Answer: 1.2 kJ

For a monatomic ideal gas U = (3/2)nRT, so ΔU = (3/2)(2.0)(8.31)(50) ≈ 1.2 kJ. Using nRΔT (0.83 kJ) or (5/2)nRΔT (2.1 kJ) are common slips.

An ideal gas is heated in a rigid sealed container (isovolumetric process). Which statement is correct?

• No work is done, so all the heat supplied increases the internal energy
• The gas does work equal to the heat supplied, so internal energy is constant
• The internal energy is constant because the volume is constant
• No heat is exchanged because the container is sealed

Answer: No work is done, so all the heat supplied increases the internal energy

With ΔV = 0 the gas does no work (W = pΔV = 0), so the first law gives Q = ΔU: every joule of heat raises the internal energy and hence the temperature.

A quantity of 500 J of thermal energy is transferred reversibly to a system kept at a constant temperature of 250 K. The entropy change of the system is:

• −2.0 J K⁻¹
• 0.50 J K⁻¹
• 1.3 × 10⁵ J K⁻¹
• +2.0 J K⁻¹

Answer: +2.0 J K⁻¹

ΔS = Q/T = 500/250 = +2.0 J K⁻¹; the entropy increases because heat enters the system. Dividing T by Q or using a negative sign are common errors.

The Carnot cycle consists of which sequence of processes?

• Two isobaric and two isovolumetric processes
• Two isothermal and two adiabatic processes
• Two isothermal and two isovolumetric processes
• Two adiabatic and two isobaric processes

Answer: Two isothermal and two adiabatic processes

The Carnot cycle is an isothermal expansion, adiabatic expansion, isothermal compression and adiabatic compression; heat is exchanged only during the two isothermal stages.

A Carnot engine has an efficiency of 40 % and rejects heat to a cold reservoir at 270 K. The temperature of the hot reservoir is:

• 450 K
• 675 K
• 378 K
• 108 K

Answer: 450 K

η = 1 − T_c/T_h, so T_h = T_c/(1 − η) = 270/0.60 = 450 K. Dividing by η instead of (1 − η) gives the wrong 675 K.

An ideal gas is compressed slowly while its temperature is kept constant. Which row gives the correct signs of the heat Q transferred to the gas, the work W done by the gas, and the change ΔU in internal energy?

• Q positive, W positive, ΔU zero
• Q negative, W negative, ΔU negative
• Q negative, W negative, ΔU zero
• Q zero, W negative, ΔU positive

Answer: Q negative, W negative, ΔU zero

Isothermal means ΔU = 0. The gas is compressed, so the work done by the gas is negative, and Q = W means the gas must release an equal amount of heat (Q negative).