Question

An insulated vessel contains 1.5 moles of argon at 2 atm. The gas initially occupies a volume of 5 L. As a result of the adiabatic expansion the pressure of the gas is reduced to 1 atm. (a) Find the volume and temperature of the final state. (b) Find the temperature of the gas in the initial state. (c) Find the work done by the gas in the process. (d) Find the change in the internal energy of the gas in the process.

Answer 1

a)

`V_f = 7.58 \, L\\T_f= 61.58 \, K`

b)

`T_0 = 81.25 \, K`

c):

`W= 367.95\,  J`

d)

`\Delta U = - 367.95\,  J`

Step-by-step explanation:

We assume ideal gas behaviour (because of low pressure) and also that Argon has ideal monoatomic gas properties, namely:

`c_v= \cfrac{3}{2}\,R\\\\c_p= \cfrac{5}{2}\,R\\\\\gamma =\cfrac{c_p}{c_v}=5/3`

For an adiabatic (constant entropy) process, we have for ideal gases that:

`P_0 {V_0} ^\gamma=P_f {V_f} ^\gamma`

We can solve a) by solving for
`V_f` as follows:

`P_0 {V_0} ^\gamma=P_f {V_f} ^\gamma\\\\V_f={\cfrac{P_0}{P_f}}^(1/ \gamma)\, V_f = 7.58 \, L`

And we use the ideal gas law to get
`T`:

`T_f = \cfrac{P_f V_f }{nR}=61.58 \, K`

For b) we just have to apply the ideal gas law again:

`PV=nRT\\\\T_0 = \cfrac{P_0 V_ 0 }{nR}=81.25 \, K`

Where we have used:

`R= 0.08205 \cfrac{L\cdot atm}{K \cdot mol}`

For c), we can either integrate, or use the fact that the internal energy of am ideal gas depends only on its Temperature by:

`\Delta U= n \,c_v  \, \Delta T`

and thus:

`\Delta T = (61.58-81.25)K=-19.67 \, K`

and so:

`\Delta U = 1.5  \,  mol \cdot \cfrac{3}{2}\,  8.314 \, \cfrac{J}{K\, mol}\, (-19.67\, K)=-367.95 \, J`

Which is the Answer to d), and we can then answer c) by the First law of thermodynamics:

`\Delta U = Q-W \\W= 367.95\,  J`

Where we have used the fact that the process is adiabatic, and thus no heat is exchanged between the system and its environment.