Question

n = 2.58 mol of Hydrogen gas is initially at T = 376 K temperature and pi = 1.88×105 Pa pressure. The gas is then reversibly and isothermally compressed until its pressure reaches pf = 8.78×105 Pa. What is the volume of the gas at the end of the compression process? What would be the temperature of the gas, if the gas was allowed to adiabatically expand back to its original pressure?

Answer #1

Use ideal gas law:

pf ∙ Vf = n ∙R∙Tf

=>

Vf = n∙ R∙Tf / pf

= 2.58 mol ∙ 8.3145 Pa∙m³∙K⁻¹∙mol⁻¹ ∙ 376 K / 8.78×10⁵ Pa

**= 918.64×10 ^{⁻5} m³
= 9.18 L**

For an ideal gas undergoing a reversible and adiabatic
process:

p∙V^γ = constant

with γ = Cp/Cv

The heat capacity ration for a diatomic ideal gas like hydrogen
(H₂) is:

γ = 7/5

Relation above can rewritten in terms of pressure and temperature
using ideal gas law:

V = n∙ R∙T/p

=>

p∙(n∙ R∙T/p)^γ = C

<=>

p^(1-γ) ∙ T^γ = C /(n∙ R)^γ = constant

<=>

T / p^[(γ - 1)/γ)] = [C /(n∙ R)^γ]^[1/γ] = constant

for γ = 7/5

T / p^(2/7) = constant

<=>

T' / pi^(2/7) = T / pf^(2/7)

=>

T' = T ∙ (pi/pf)^(2/7)

= 376 K ∙ (1.88×10⁵ Pa / 8.78×10⁵ Pa)^(2/7)

**= 242 K**

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