Answer:
W= 2.79 atm.L
ΔU= -2.79 atm.L
Explanation:
Given that
T₁ = 61⁰C
T₂=10⁰C
n= 0.44 moles
We know that specific heat capacity ratio for Ar
γ = 1.66
R=gas constant = 0.0821 atm.L/K.mol
The work done given as
[tex]W=nR\times \dfrac{T_1-T_2}{\gamma - 1}[/tex]
[tex]W=0.44\times 0.0821\times \dfrac{61-10}{1.66- 1}[/tex]
W= 2.79 atm.L
Given that process is adiabatic that is why heat transfer is zero.
From first law
Q= W + ΔU
ΔU=Change in the internal energy
Q=Heat transfer
W= Work transfer
0 = 2.79 + ΔU
ΔU= -2.79 atm.L
