Question

One mole of the gas Ar expands through a reversible adiabatic process, from a volume of 1 L and a temperature of 300 K, to a volume of 5 L.

A) what is the final temperature of the gas?

B) How much work has the expansion carried out?

C) What is the change in heat?

Assume this is a mono-atomic ideal gas.

Note by asker: as the gas is mono-atomic, c_p=(5/2)*R, c_v=(3/2)*R

Answer 1

a)Adiabatic coefficient k= c_p/c_v = 5/3

In adiabatic process , T V^K-1 remains constant (where T is temperature and V is volume)

So, 300*1^2/3 = T*5^2/3 where T is the final temperature

So,final temperature T= 300/5^2/3 =102.6 K

b)initial volume=1 L= 0.001 m³ , initial temperature = 300 K

Also, number of moles= 1

So, using ideal gas equation, PV=nRT=>P=nRT/V=8.314*300/0.001=2494.2 *10³ pa

So, initial pressure= 2494.2 Kpa

Similarly,final pressure=8.314*102.6/0.005=170.6 *10³ pa

So, work out= (P₂V₂-P₁V₁)/(1-k)=(170.6*5-2494.2*1)/(1-5/3)=2461.8 J

c)No heat is supplied in adiabatic process, So, heat=0 J

However, change in internal energy= - work out= -2461.8 J