n = 3.55 mol of Hydrogen gas is initially at T = 323 K temperature and pi = 2.76

Question

n = 3.55 mol of Hydrogen gas is initially at T = 323 K temperature and pi = 2.76×105 Pa pressure. The gas is then reversibly and isothermally compressed until its pressure reaches pf = 7.03×105 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? Please Show ALL work, Thank you!

Explanation / Answer

no of mole of Hydrogen, n=3.55

Temperature T=323 K

Pressure P1=2.76*10^5 Pa

Pressure P2=7.03*10^5 Pa

a)

by using ideal gas equation,

P*v=n*R*T

P2*v2=n*R*T

7.03*10^5*v2=3.55*8.314*323

====>

v2=1.36*10^-2 m^3

volume of the gas at the end of the compression is, V=1.36*10^-2 m^3

b)

work done dW=n*R*T*ln(P1/P2)

=3.55*8.314*323*ln((2.76*10^5)/(7.03*10^5))

=-8913.2 J

workdone on the gas, dw=8913.2 J

c)

by using energy relation,

dQ=dU+dW

here, in case of isothermal process, dU=0

and

dQ=dW

hence, heat emitted, dQ=8913.2 J

d)

entropy , dS=dQ/T

dS=8913.2/323

dS=27.6 J/K

e)

in case of adiabatic process,

T^gama/(P^gama-1) =constant

====>

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

here,

for gama=7/5 ( hydrogen is a dia-atomic gas)

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

(323/T2)^(7/5) = ((2.76*10^5)/(7.03*10^5))^((7/5)-1)

===> T2=421.91 K

temperature T2=421.9 K or T2=422 K

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