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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

  1. q = 0
  2. T2=T1
  3. p2V2=p1V1
  4. p2Vγ2=p1Vγ1


Solution

The correct options are
A q = 0
B T2=T1
C p2V2=p1V1

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

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