N = 2.89 mol of Hydrogen gas is initially at T = 387 K temperature and p_1 = 2.5

Question

N = 2.89 mol of Hydrogen gas is initially at T = 387 K temperature and p_1 = 2.59 times 10^5 Pa pressure. The gas is then reversibly and isothermally compressed until its pressure reaches p_f = 7.39 times 10^5 Pa. What is the volume of the gas at the end of the compression process? How much work did the external force perform? How much heat did the gas emit? How much entropy did the gas emit? What would be the temperature of the gas, if the gas was allowed to adiabatically expand back to its original pressure?

Explanation / Answer

For an isothermal process

PV = nRT = constant

So PiVi = nRTi = PfVf

Vf = nRTi/Pf

n = 2.89 moles
R = 8.31 J/(mol.°K)
Ti = 387 °K
Pf = 7.39 x 10^5 Pa
Vf = 2.89*8.31*387 / (2.59 * 10^5) m^3

Vf = 3.588 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 = 2.89*8.31*387 ln(7.39 * 10^5/ (2.59*10^5)) J

W = 9744.639 J <===

E = Q + W

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

Thus Q = - W = - 9744.639 J is the heat emitted <===

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

For an isothermal process, Tf = Ti and

S = nR ln(Vf/Vi)

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

S = nR ln(Pi/Pf) = 2.89*8.31 ln(2.59 * 10^5/ (7.39*10^5)) J/°K

S = -25.1789 J/°K <=== is the change of entropy

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

So P1/P2 = (V2/V1)^....(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))^

P1/P2 = (P1/P2)^ (T2/T1)^

So (P1/P2)^(1-) = (T2/T1)^

ln(T2/T1) = (1-) ln(P1/P2)

lnT2 - lnT1 = (1-)/ * ln(P1/P2)

lnT2 = lnT1 + (1-)/ * ln(P1/P2)

T1 = 387 °K
= 1.384 for hydrogen
P1 = 7.39 x 10^5 Pa
P2 = 2.59 x 10^5 Pa so

lnT2 = ln387 + (1-1.384)/1.384 ln(7.39 * 10^5/ (2.59*10^5))

lnT2 = 5.9584 - 0.2908904

T2 = 289.3131 °K <=== is the final temperature

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