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Question

The initial internal energy of the gas in container 'A', if the containers were at room temperature 300K initially-
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A
1406.25 cal
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B
1000 cal
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C
2812.5 cal
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D
none of these
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Solution

The correct option is D 2812.5 cal
Since the temperature rise and heat supplied to both the container is same.

Q=nACVAΔT=nBCVBΔT nACVA=nBCVB (1)

also, ΔPAV=nARΔT (2)

and, ΔPBV=nBRΔT (3)

Dividing (2) by (3),

nAnB=ΔPAΔPB=2.51.5=53

nA=5nB3

From equation (1)
CVBCVA=53=(5R/2)(3R/2)

CVB=5R/2, so B is diatomic.

CVA=3R/2, so A is Mono-atomic.

Now,
nA=12540,CVA=3R2

Internal energy is Ui=nACVAT

Ui=125×3×240×2×300=2812.5 cal

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