Question

n = 4.13 mol of Hydrogen gas is initially at T = 367.0 K temperature and p_i = 1.85×10⁵ Pa pressure. The gas is then reversibly and isothermally compressed until its pressure reaches p_f = 6.59×10⁵ Pa. What is the volume of the gas at the end of the compression process?

A. How much work did the external force perform?

B. How much heat did the gas emit?

C. How much entropy did the gas emit?

D. What would be the temperature of the gas, if the gas was allowed to adiabatically expand back to its original pressure?

Answer 1

A) workdone by extenral agent, W = n*R*T*ln(Pf/Pi)

= 4.13*8.314*367*ln(6.59/1.85)

= 1.60*10^4 J <<<<<<<<<<-------------------Answer

B) In Isothermal process, delta_U = 0

use first law of thermodynamics,

delta_Q = W + delta_U

delta_Q = W

heat emitted, delta_Q = W

= 1.60*10^4 J <<<<<<<<<<-------------------Answer

C) entropy emited by the gas, delta_S = delta_Q/T

= 1.60*10^4/367

= 43.6 J/K <<<<<<<<<<-------------------Answer

d)

we know for hydrogen gas, gamma = 1.4

here, Pi = 6.59*10^5 pa,
Pf = 1.85*10^5 Pa
Ti = 367 K
Tf = ?

in adiabatic process, P^(1-gamnma)*T^gamma = constant

Pf^(1-gamma)*Tf^gamma = Pi^(1-gamma)*Ti^gamma

Tf^gamma = (Pi/Pf)^(1 - gamma)*Ti^gamma

= (Pi/Pf)^(1/gamma - 1)*Ti

= (6.59/1.85)^(1/1.4 - 1)*367

= 255 K <<<<<<<<<<-------------------Answer