n = 2.92 mol of hydrogen gas is initially at t ... Question: N = 2.92 mol of Hyd
ID: 1641183 • Letter: N
Question
n = 2.92 mol of hydrogen gas is initially at t ... Question: N = 2.92 mol of Hydrogen gas is initially at T = 3... n = 2.92 mol of Hydrogen gas is initially at T = 343 K temperature and pi = 1.68×105 Pa pressure. The gas is then reversibly and isothermally compressed until its pressure reaches pf = 7.59×105 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 write out all work to questions. .
The following answers are incorrect:
a).22 Vi
b) -12554 J
c)12554 J
d)36.6 J/K
e) 187.2 K
PLEASE GIVE DIFFERENT VALUES THESE ARE NOT CORRECT!!!!!!
Explanation / Answer
1.
Using ideal gas law:
PV = nRT
V = nRT/P
V = 2.92*8.314*343/(1.68*10^5) = 49.56*10^-3 m^3
V = 49.56 Litres
2.
Work done is given by:
W = -nRT*ln (Vf/Vi) = nRT*ln (Pf/Pi)
W = 2.92*8.314*343*ln (7.59/1.68) = +12557.39 J
3.
dU = Q + W
since in isothermal process, dU = 0
Q = -W
Q = -12557.39 J
4.
dS = dQ/T
dS = -12557.39/343 = -36.61 J/K
5.
In adiabatic process
PV^y = Constant
OR
P^(1 - y)*T^y = C
T*P^((1-y)/y) = C
for Hydrogen gas, y = 7/5
T*P^((1 - 7/5)/7/5) = T*P^(-2/7) = C
T2/T1 = (P2/P1)^(2/7)
T2 = T1*(P2/P1)^(2/7)
T2 = 343*(1.68/7.59)^(2/7) = 222.93 K
Let me know if you have any doubt.
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