Answer:
a)
[tex]V_f = 7.58 \, L\\T_f= 61.58 \, K[/tex]
b)
[tex]T_0 = 81.25 \, K[/tex]
c):
[tex]W= 367.95\, J[/tex]
d)
[tex]\Delta U = - 367.95\, J[/tex]
Explanation:
We assume ideal gas behaviour (because of low pressure) and also that Argon has ideal monoatomic gas properties, namely:
[tex]c_v= \cfrac{3}{2}\,R\\\\c_p= \cfrac{5}{2}\,R\\\\\gamma =\cfrac{c_p}{c_v}=5/3[/tex]
For an adiabatic (constant entropy) process, we have for ideal gases that:
[tex]P_0 {V_0} ^\gamma=P_f {V_f} ^\gamma[/tex]
We can solve a) by solving for [tex]V_f[/tex] as follows:
[tex]P_0 {V_0} ^\gamma=P_f {V_f} ^\gamma\\\\V_f={\cfrac{P_0}{P_f}}^{1/ \gamma}\, V_f = 7.58 \, L[/tex]
And we use the ideal gas law to get [tex]T[/tex]:
[tex]T_f = \cfrac{P_f V_f }{nR}=61.58 \, K[/tex]
For b) we just have to apply the ideal gas law again:
[tex]PV=nRT\\\\T_0 = \cfrac{P_0 V_ 0 }{nR}=81.25 \, K[/tex]
Where we have used:
[tex]R= 0.08205 \cfrac{L\cdot atm}{K \cdot mol}[/tex]
For c), we can either integrate, or use the fact that the internal energy of am ideal gas depends only on its Temperature by:
[tex]\Delta U= n \,c_v \, \Delta T[/tex]
and thus:
[tex]\Delta T = (61.58-81.25)K=-19.67 \, K[/tex]
and so:
[tex]\Delta U = 1.5 \, mol \cdot \cfrac{3}{2}\, 8.314 \, \cfrac{J}{K\, mol}\, (-19.67\, K)=-367.95 \, J[/tex]
Which is the Answer to d), and we can then answer c) by the First law of thermodynamics:
[tex]\Delta U = Q-W \\W= 367.95\, J[/tex]
Where we have used the fact that the process is adiabatic, and thus no heat is exchanged between the system and its environment.
