n = 2.95 mol of Hydrogen gas is initially at T = 320 K temperature and p i = 2.1
ID: 1486434 • Letter: N
Question
n = 2.95 mol of Hydrogen gas is initially at T = 320 K temperature and pi = 2.14×105 Pa pressure. The gas is then reversibly and isothermally compressed until its pressure reaches pf = 6.51×105 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?
C. How much heat did the gas emit?
D. 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
from ideal gas equation
P1*v1 = n*R*T1
2.14*10^5 *V1 = 2.95*8.314*320
v1 = 0.0367 m^3
In isothermal process
p1*v1 = p2*v2
2.14*10^5*0.0367 = 6.15*10^5*v2
v2 = 0.01278 m^3 <<-----answer
B)
In isothermal process
W = 2.303*n*R*T*log(P1/P2)
W = 2.303*2.95*8.314*320*log(2.14/6.51)
W = -8733.2 J <<-------answer
C)
from first law of thermodynamics
Q = U + W
U = n*Cv*dT = 0
Q = 8733.2 J
D)
D = Q/T = 8733/2/320 = 13.54 J/K
IN adiabatic process
P1^(1-gamma)*T1^(gamma) = P2^(1-gamma)*T2^(gamma)
gamma = ratio of specific heat = 1.4
(2.14*10^5)^(1.4-1)*320^1.4 = (6.51*10^5)^(1.4-1)*T2^1.4
T2 = 232.86 K
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