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
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