thermodynamic transformation

Question

One mole of monatomic gas expands adiabatically until reaching a volume three times the initial one. Assuming that the initial temperature is 100 °C, what will be the work done during the thermodynamic transformation?

Answer 1

Monoatomic gas, C_V = 3R/2, C_P = 5R/2, γ = 5/3 = 1.67

Adiabatic: 0 = ΔU + W, W = - ΔU = -nC_V(T_f - T_i) (1)

On the other hand, for an adiabatic process: T_f/T_i = (V_i/V_f)^γ-1 = (1/3)^0.67 = 0.48

T_i = 373 K, T_f = 48.1 K

Back in eq. (1): W = -1 · 3/2 · 8.31 ·(48.1 - 373) J = 4050 J

This is the work done by the gas, and not on the gas. It is positive and that is correct: the gas is expanding, so it does work.

Answer 2

Monoatomic gas, C_V = 3R/2, C_P = 5R/2, γ = 5/3 = 1.67

Adiabatic: 0 = ΔU + W, W = - ΔU = -nC_V(T_f - T_i) (1)

On the other hand, for an adiabatic process: T_f/T_i = (V_i/V_f)^γ-1 = (1/3)^0.67 = 0.48

T_i = 373 K, T_f = 48.1 K

Back in eq. (1): W = -1 · 3/2 · 8.31 ·(48.1 - 373) J = 4050 J

This is the work done by the gas, and not on the gas. It is positive and that is correct: the gas is expanding, so it does work.