Question

Three moles of an ideal monatomic gas expand at a constant pressure of 2.40 atm ; the volume of the gas changes from 3.20×10^-2 m³ to 4.50×10⁻² m³ .

a. Calculate the initial temperature of the gas.

b. Calculate the final temperature of the gas.

c. Calculate the amount of work the gas does in expanding.

d. Calculate the amount of heat added to the gas.

e. Calculate the change in internal energy of the gas.

Answer 1

  part A
Use ideal gas law
P₁∙V₁ = n∙R∙T₁
=>
T₁ = P₁∙V₁ / (n∙R₁)
= 2.4 ∙ 101325 Pa ∙ 3.2×10⁻² m³ / ( 3 mol ∙ 8.314472 Pa∙m²∙K⁻¹∙mol⁻¹)
= 311.97 K


part B
From ideal gas follows that
V/T = n∙R/P = constant
for a constant pressure process.
Hence,
V₁/T₁ = V₂/T₂
=>
T₂ = T₁ ∙ ( V₂/V₁)
= 311.97 K ∙ ( 4.5×10⁻² m³ / 3.2×10⁻² m³)
= 438.70 K


part C
Work done by the gas is given by the integral:
W = ∫ P dV from initial to final volume
For a constant pressure process this simplifies to:
W = P ∙∫ dV = P∙∆V
Hence,
W = P∙(V₂ - V₁)
= 2.4 ∙ 101325 Pa ∙ (4.5×10⁻² m³∙ - 3.2×10⁻² m³)
= 3161.34 Pa∙m³
= 3161.34 J


part D
Heat transferred in a constant pressure process equals the change in enthalpy:
Q = ∆H
Change in enthalpy for an ideal gas is given by:
∆H = n∙Cp∙∆T
The molar heat capacity at constant pressure for a monatomic ideal gas is:
Cp = (5/2)∙R
Hence,
Q = (5/2)∙n∙R∙(T₂ - T₁)
= (5/2) ∙ 3 mol ∙ 8.314472 J∙K⁻¹∙mol⁻¹ ∙ (438.70 K - 311.97 K)
= 7902.24 J


part E
Change in internal energy equals heat added to minus work done by the gas:
∆U = Q - W
= 7902.24 J - 3161.34 J
= 4740.90 J