Question

Pressure-temperature relationship for an ideal gas undergoing adiabatic change is:

$\gamma =C_p/C_v$

• A. ${PT}^{\gamma }=$ constant
• B. ${PT}^{-1+\gamma }=$ constant
• C. ${PT}^{-1}{T}^{\gamma }=$ constant
• D. $P^{1-\gamma }{T}^{\gamma }=$ constant

Answer 1

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

From the first law of thermodynamics, $dU=dQ+dW$

But for adiabatic process, $dQ=0$

$dU=dW$

$dU=n{C}_{v}dT$

$dW=-PdV$

$n{C}_{v}dT=-PdV\Rightarrow n{C}_{v}dT+PdV=0$

For one mole of gas, $n=1$ $C_{v}dT+PdV=0$

For ideal gas, $PV=RT$

$P=\frac{RT}{V}$

$C_{v}dT+\frac{RT}{V}dV=0$

$\Rightarrow C_v\frac{dT}{T}+R\frac{dV}{V}=0$ $[C_p-C_v=R]$

$\Rightarrow C_v\frac{dT}{T}+(C_p-C_v)\frac{dV}{V}=0$ $\gamma =\frac{C_p}{C_v}$

$\Rightarrow \frac{dT}{T}+(\frac{C_p}{C_v}-1)\frac{dV}{V}=0$

$\Rightarrow \frac{dT}{T}+(\gamma -1)\frac{dV}{V}=0$

$\int \frac{1}{T}dT+(\gamma -1)\int \frac{1V}{d}V=0$

On integration,

$lnT+(\gamma -1)lnV=lnC$

$\Rightarrow lnT{V}^{\gamma -1}=lnC$

$\Rightarrow T{V}^{\gamma -1}=C$

$T{\left(\frac{RT}{P}\right)}^{1-\gamma }=C$ [$PV=RT$; \ $V=\frac{RT}{P}$]

$\Rightarrow T^{\gamma }{P}^{(1-\gamma )}=constant$