Question

An ideal monatomic gas is confined to a rigid container. When heat is added reversibly to the gas, its temperature changes from T₁ to T₂. (a) How much heat is added? (b) What is the change in entropy of the gas?

a) nCᵥ(T₂ - T₁), nCᵥ ln(T₂/T₁)
b) nCᵥ(T₁ - T₂), nCᵥ ln(T₂/T₁)
c) nCᵥ(T₂ - T₁), nCᵥ(T₁ - T₂)
d) nCᵥ(T₁ - T₂), nCᵥ(T₂ - T₁)

Answer 1

For a monatomic ideal gas in a rigid container, the heat added during a reversible process is nCv(T2 - T1), and the change in entropy is nCv ln(T2/T1), representing option (c).

Step-by-step explanation:

The ideal monatomic gas in a rigid container undergoing a reversible heat addition process gives rise to two main outcomes: the amount of heat added and the change in entropy of the gas.

(a) The amount of heat added to the system can be calculated using the specific heat at constant volume, Cv, and the change in temperature. Since the process is at constant volume and the gas is ideal, the amount of heat added, Q, is given by:

Q = nCv(T2 - T1)

(b) The change in entropy ΔS at constant volume for a reversible process is given by:

ΔS = nCvln(T2/T1)

Therefore, the correct answers are: (a) nCv(T2 - T1), and (b) nCvln(T2/T1), which corresponds to option (c).