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n = 2.42 mol of Hydrogen gas is initially at T = 319 K temperature and p i = 2.1

ID: 1613186 • Letter: N

Question

n = 2.42 mol of Hydrogen gas is initially at T = 319 K temperature and pi = 2.18×105 Pa pressure. The gas is then reversibly and isothermally compressed until its pressure reaches pf = 9.62×105 Pa.

A) What is the volume of the gas at the end of the compression process?

B) How much work did the external force perform?
9.52×103 J

C) How much heat did the gas emit?
9.52×103 J

D) How much entropy did the gas emit?
2.99×101 J/K

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

PLEASE ANSWER A AND E. THE ANSWERS ARE NOT .2266 Vi for A or 176.15 K for E.

Explanation / Answer

given

n = 2.42 mol

T = 319 K

Pi = 2.18*10^5 pa

Pf = 9.62*10^5 pa

A) Vi = ?

use Ideal Gas equation

Pi*Vi = n*R*Ti

Vi = n*R*Ti/Pi

= 2.42*8.314*319/(2.18*10^5)

= 0.0294 m^3 <<<<<<<<<<<<<---------------------Answer

E) Vf = n*R*Tf/Pf

= 2.42*8.314*319/(9.62*10^5)

= 0.00667 m^3

now

let T1 = 319 k

P1 = 9.62*10^5 m^3

P1 = 2.18*10^5 m^3

T2 = ?

for hydrogen, gamma = 1.4

P^(1-gamma)*T^gamma = constant.

P2^(1-gamma)*T2^gamma = P1^(1-gamma)*T1^gamma

T2^gamma = T1^gamma*(P1/P2)^(1-gamma)

T2 = T1*(P1/P2)^(1/gamma - 1)

= 319*(9.62/2.18)^(1/1.4 - 1)

= 209 K <<<<<<<<<<<<<---------------------Answer