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Question

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


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

The correct option is D $${ P }^{ 1-\gamma }{ T }^{ \gamma }=$$ constant
From 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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