Question

An ideal gas in a thermally insulated vessel at internal pressure `=P_(1)`, volume `=V_(1)` and absolute temperature `=T_(1)` expands irreversibly against zero external pressure, as shown in the diagram. The final internal pressure, volume and absolute temperature of the gas are `P_(2), V_(2)` and `T_(2)` respectively. For this expansion.

A. `q=0`
B. `T_(2)=T_(1)`
C. `P_(2)V_(2)=P_(t)V_(1)`
D. `P_(2)V_(2)^(gamma)=P_(1)V_(1)^(gamma)`

Answer 1

Correct Answer - A::B::C
Since the vessel is thermally insulated so
`q=0`
`p_(ext)=0`, so `w=0`
so `Delta U=0` (ideal gas)
Hence `Delta T=0`
`rArr Delta T=0 " "rArrT_(2)=T_(1) " "rArrP_(2)V_(2)=P_(1)V_(1)`
The process is however adiabatic irriversible.
So we cannot applly `P_(2)V_(2)^(gamma)=P_(1)V_(1)`
Hence ans is `(A),(B),(C )`