Question

# Pressure-temperature relationship for an ideal gas undergoing adiabatic change is: $$\gamma ={ C }_{ p }/{ C }_{ v }$$

A
PTγ= constant
B
PT1+γ= constant
C
PT1Tγ= constant
D
P1γTγ= constant

Solution

## The correct option is D $${ P }^{ 1-\gamma }{ T }^{ \gamma }=$$ constantFrom the first law of thermodynamics, $$dU=dQ+dW$$But for adiabatic process, $$\boxed {dQ=0}$$$$dU=dW$$$$dU=nC_vdT$$$$dW=-PdV$$$$nC_vdT=-PdV \Rightarrow nC_vdT+PdV=0$$For one mole of gas, $$n=1$$ $$\boxed {C_vdT+PdV=0}$$For ideal gas, $$PV=RT$$$$\boxed {P= \cfrac {RT}V}$$$$C_vdT + \cfrac {RT}V dV=0$$$$\Rightarrow C_v \cfrac {dT}T + R \cfrac {dV}V=0$$ $$[C_p - C_v=R]$$$$\Rightarrow C_v \cfrac {dT}T + (C_p - C_v) \cfrac {dV}V=0$$ $${\gamma = \cfrac {C_p}{C_v}}$$$$\Rightarrow \cfrac {dT}T + \left (\cfrac {C_p}{C_v}-1 \right) \cfrac {dV}V=0$$$$\Rightarrow \cfrac {dT}T + (\gamma -1) \cfrac {dV}V=0$$$$\int {\cfrac 1T dT} + (\gamma - 1) \int {\cfrac {1V}dV=0}$$On integration, $$ln \ T + (\gamma -1) ln \ V= ln \ C$$$$\Rightarrow ln \ TV^{\gamma -1}= ln \ C$$$$\Rightarrow TV^{\gamma -1}=C$$$$T \left (\cfrac {RT}P \right)^{1- \gamma}=C$$ [$$PV=RT$$; \ $$V= \cfrac {RT}P$$]$$\Rightarrow \boxed {T^{\gamma}P^{(1- \gamma)}=constant}$$Chemistry

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