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 m^{3} , 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 *10^{3} pa

So, initial pressure= 2494.2 Kpa

Similarly,final pressure=8.314*102.6/0.005=170.6 *10^{3}
pa

So, work out=
(P_{2}V_{2}-P_{1}V_{1})/(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

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