Question

n = 2.94 mol of Hydrogen gas is initially at T = 381 K temperature and p_i = 2.24×10⁵ Pa pressure. The gas is then reversibly and isothermally compressed until its pressure reaches p_f = 7.63×10⁵ 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

For an isothermal process

PV = nRT = constant

So PiVi = nRTi = PfVf

Vf = nRTi/Pf

n = 2.94 moles
R = 8.31 J/(mol.°K)
Ti = 381 °K
Pf =7.63 x 10^5 Pa
Vf = 2.94*8.31*381 / (2.24 * 10^5) m^3

Vf =4.155 x 10^(-2) m^3 <===

For an isothermal process, the work is

W = nRT ln(Vi/Vf)
Since PiVi = PfVf, we have Vi/Vf = Pf/Pi and

W = 2.94*8.31*381 ln(7.63 * 10^5/ (2.24*10^5)) J

W = 11408.4J <===

ΔE = Q + W

For an isothermal process, the change in internal energy is zero : ΔE = 0

Thus Q = - W = - 11408.4 J is the heat emitted <===

ΔS = nCv ln(Tf/Ti) + nR ln(Vf/Vi) for the entropy change of an ideal gas

For an isothermal process, Tf = Ti and

ΔS = nR ln(Vf/Vi)

From PiVi = PfVf, we have Vf/Vi = Pi/Pf so

ΔS = nR ln(Pi/Pf) = 2.94*8.31 ln(2.24 * 10^5/ (7.63*10^5)) J/°K

ΔS = -29.94 J/°K <=== is the change of entropy

For an adiabatic process, PV^γ = constant where γ = 1.384 for hydrogen

So P1/P2 = (V2/V1)^γ....(i)

Note: V1, P1 are the initial values and V2, P2 the final values of the volume
and pressure in this question.

We use the gas law PV = nRT to replace V2/V1

P1V1 = nRT1 and P2V2 = nRT2 which gives
V2/V1 = (nRT2/P2)/(nRT1/P1)
or V2/V1 = P1 T2/ (P2 T1). Substitute this in eqn (i)

P1/P2 = (P1 T2/ (P2 T1))^γ

P1/P2 = (P1/P2)^γ (T2/T1)^γ

So (P1/P2)^(1-γ) = (T2/T1)^γ

γ ln(T2/T1) = (1-γ) ln(P1/P2)

lnT2 - lnT1 = (1-γ)/γ * ln(P1/P2)

lnT2 = lnT1 + (1-γ)/γ * ln(P1/P2)

T1 = 381 °K
γ = 1.384 for hydrogen
P1 = 7.63 x 10^5 Pa
P2 = 2.24 x 10^5 Pa

so

lnT2 = ln381 + (1-1.384)/1.384 ln(7.63 * 10^5/ (2.24*10^5))

lnT2 = -.785

T2 = 273.4 °K <=== is the final temperature