sample of 3 mole of an ideal gas at 200Kelvin and 2ATM is compressed riversibly and adibaticaly until the temperature reaches 250 Kelvin given that molar heat capacity is 27.5J K^-1 mol^-1 at constant volume
Answers
Answer:
The initial values are n=3 mole,T
1
=200K,P
1
=2.0 atm,C
v
=27.5 JK
−1
mol
−1
The final values (after compression) are T
2
=250 K,P
2
=?
The heat capacity at constant pressure is C
p
=27.5+8.314 JK
−1
mol
−1
=35.814 JK
−1
mol
−1
Also γ=
C
v
C
p
=
27.5
35.814
=1.30
But p
1
1−γ
⋅T
1
γ
=P
2
1−γ
⋅T
2
γ
or P
2
1−γ
=P
1
1−γ
(
T
2
T
1
)
γ
Substitute values in the above expression.
(P
2
)
−0.3
=(2)
−0.3
×(
250
200
)
1.30
Thus, P
2
=5.26atm
The initial volume is V
1
=
P
1
nRT
=
2
3×0.0821×200
=24.63litre
Also, P
1
V
1
γ
=P
2
V
2
γ
or [
V
1
V
2] γ = P 2bP 1
Substitute values in the above expression.
[ 24.63V 2]
1.3 = 5.22
Hence, V
2
=11.81 litre
The work done is =
γ−1
nR
[T 2−T 1 ]= 0.3
3×8.314
×[250−200]=+4157J=+4.157 kJ
Since the process is adiabatic, no heat is transferred. q=0
Internal energy decreases as work is done. Hence, Δ U=w=4.157 kJ
Also, the expression for the enthalpy change is ΔH=n×C
p
×ΔT
Substitute values in the above expression.
ΔH=3×35.814×50=5372.1 J=5.372 kJ
Explanation:
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