Question

2 litres of an ideal gas at a pressure of 10atm expands isothermally into vacuum until its total volume is 10 litres. How much work is done? How much heat is absorbed? Consider the same expansion but this time the constant external pressure of 1 ATM?

Answer 1

In the case of expansion of any gas, work is done by the system.
Now we know that
w = -pext ΔV
= -pext (Vf-Vi)

where pext​ = external pressure of the system = 0

(The given pressure of 10 atm in the question, is the pressure of the gas itself. But we need external pressure in the expression. The gas is expanding into vacuum which has no pressure. Thus external pressure = 0)
Vf = 10 L
Vi = 2 L

Thus work done by the system will be:
w = -0×(10-2)
= 0
Thus no work is done.
Also, heat absorbed (q) is a consequence of gaining temperature, but the system is working at constant temperature (isothermal condition).
q = -w
So there will be no heat absorbed.


pext = 1 atm
Vf = 10 L
Vi = 2 L
So w = -1 atm ×(10-2) L = - 8 atm L
Also, during isothermal expansion,
q = - w = 8 atm L