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Question

An ideal gas in thermally insulated vessel at internal pressure =p1,volume=V1 and absolute temperature =T1 expands irreversibly against zero external pressure, as shown in the diagram.
The final internal pressure, volume and absolute temperature of the gas are p2,V2 and T2, respectively. For this expansion

A
q = 0
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B
T2=T1
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C
p2V2=p1V1
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D
p2Vγ2=p1Vγ1
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Solution

The correct option is C p2V2=p1V1
PLAN This problem includes concept of isothermal adiabatic irreversible expansion.
Process is adiabatic because of the use of thermal insolution therefodre, q = 0
Pext=0
w=Pext.V=0×V=0
Internal energy can be written as \triangle U = q + W = 0
The change in internal energy fo an ideal gas depends only on temperature and change in internal energy ( U) = 0 therefore, T = 0 hence, process is isothermal and T2=T1 and p2V2=p1V11
(d) p2Vγ2=(d)p1Vγ1 is incorrect, it is valid for adiabatic reversible process.
Hence, only (a), (b) and (c) are correct choices.

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