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Question

Two moles of an ideal monoatomic gas occupies a volume V at 27oC. The gas expands adiabatically to a volume 2V.

Calculate (a) the final temperature of the gas and (b) change in its internal energy.

Options

A.

(a) 195 K (b) - 2.7 kJ

B.

(a) 189 K (b) - 2.7 kJ

C.

(a) 195 K (b) 2.7 kJ

D.

(a) 189 K (b) 2.7 kJ

Solution

In an adiabatic process
PVν = constant
and, PV = nRT, gives
⇒ Vγ-1∝ 1T
V1V2γ-1=T2T1
T2 = T1 V1V2γ-1
monoatomic gas : γ =53
⇒T2= (300 K) V2V53-1
= 189 K (final temperature)
change in internal energy
∆U=nf2R  ∆T
=232253(-111)
= - 2.7 kJ

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