T1=273+27=300 KT2=273+927=1200 K Gas equation for adiabatic process pVγ= constant p(pT)γ= constant (∵pV=RT)∴p1p2=(T1T2)γ−1γp2=p1(T1T2)γ−1γp2=2(3001200)1.4−11.4p2=256 atm
A mass of diatomic gas (γ=1.4) at a pressure of 2 atm is compressed adiabatically so that its temperature rise from 27∘C to 927∘C. The pressure of the gas is final state is
Held on 30 Apr 2011 · Verified 9 Jul 2026.
28 atm
68.7 atm
256 atm
8 atm
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In an adiabatic process:
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