One mole of an ideal gas at an initial temperature of T K does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas will be:
Text Solution
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In an adiabatic process, there is no heat transfer into or out of a system i.e., Q = 0.
In an adiabatic process Q = 0
So, from Ist law of thermodynamics.
W = – Δ U = – nC V Δ T
= 
= 
……(i)
Here: W = 6R J, n = 1 mol,
R = 8.31 J/mol-K, γ = 
, Ti = TK
Substituting given values in Eq. (i), we get
∴ 6R = 
⇒ 6R = 
⇒ T – T f = 4
∴ T f = (T – 4) K
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