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Question

A rigid and insulated tank of 3 m3 volume is divided into two components. One compartment of volume of 2 m3 contains an ideal gas at 0.8314 MPa and 400 K and while the second compartment of volume 1 m3 contains the same gas 8.314 MPa and 500 k. If the partition between the two compartments is ruptured, the final temperature of the gas is:

A
420 K
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B
450 K
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C
480 K
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D
None of these
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Solution

The correct option is A 420 K
First compartment
P1=0.8314 M Pa V1=2 xm2 T1=400 K
Second compartment
P2=8.314 M Pa V2=1 m3 T2=500 K
n1=P1V1RT1=0.8314×106×28.314×400=500
n2=P2V2RT2=8.314×1×1068.314×500=200
After Ruptewe of compartment
n=500+200=700 V=2 P1=3 m3
P=0.8314 M Pq
T=PVnR=3×8.314×106700×8.314
=30007
420 K

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