A certain amount of an ideal monatomic gas is heated from a temperature of T1 = 300 K to a temperature of T2

A certain amount of an ideal monatomic gas is heated from a temperature of T1 = 300 K to a temperature of T2 = 400 K. In this case, the volume of the gas varied in direct proportion to its absolute temperature (V = αT, α = const). The initial volume of gas V1 = 3 liters, the pressure measured at the end of the process turned out to be P2 = 1 atm. What work did the gas do in this process? How much heat was supplied to the gas?

1) A certain amount of an ideal monatomic gas is heated from the temperature T₁ = 300 K to the temperature T₂ = 400 K, while the volume of the gas changes in direct proportion to its absolute temperature V = α ∙ T, α = const. Since the initial gas volume is V₁ = 3 l = 0.003 m³, then V₁ = α ∙ T₁, which means that α ∙ 300 = 0.003; α = 0.00001. Then V₂ = α ∙ T₂; V₂ = 0.00001 ∙ 400; V₂ = 0.004 m³.

2)  From the condition of the problem it is known that the pressure measured at the end of the process turned out to be equal to p 1 = 1 atm = 10⁵ Pa, therefore, from the Mendeleev-Clapeyron equation (p₁ ∙ V₁) / T₁ = (p₂ ∙ V₂) / T₂ we find p₁ = (p₂ ∙ V₂ ∙ T₁) / (V₁ ∙ T₂) or p₁ = (10⁵ ∙ 0.004 ∙ 300) / (0.003 ∙ 400); p₁ = 10⁵ Pa. The process turned out to be isobaric, since p₁ = p₂, then the gas in this process did work A = p ∙ ΔV or A = 10 ∙ (0.004 – 0.003); A = 100 J.

3) The amount of heat was supplied to the gas Q = ΔU + A, where the change in internal energy ΔU = 1.5 ∙ ν ∙ R ∙ ΔT or ΔU = 1.5 ∙ ΔT ∙ (p₂ ∙ V₂) / T₂, since p₂ ∙ V₂ = ν ∙ R ∙ T₂. We get: Q = (1.5 ∙ ΔT ∙ (p₂ ∙ V₂) / T₂) + A or Q = (1.5 ∙ (400 – 300) ∙ (10⁵ ∙ 0.004) / 400) + 100; Q = 150 J.

Answer: work 100 J, amount of heat 150 J.