The adiabatic constant γ of a gas is 3/2. Its initial temperature is 0°C and its volume is reduced adiabatically to one-fourth. The new temperature is:
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The adiabatic constant γ of a gas is 3/2. Its initial temperature is 0°C and its volume is reduced adiabatically to one-fourth. The new temperature is:

G1 Physics · Accepted Answer

For an adiabatic process T V ( γ − 1 ) = c o n s t a n t . V^(γ-1)=constant. V ( γ − 1 ) = co n s t an t . With γ-1=1/2 and V₂=V₁/4: T₂=T₁(V₁/V₂)^(1/2)=273×√4=546 K.