Answer to Question #179162 in Molecular Physics | Thermodynamics for Anand

Question

One gram mole of a monoatomic perfect gas λ =5/3 at 27° C is adiabatically compressed from an initial pressure of 1 atm to final pressure of 50 atm. Calculate 

the resulting final temperature.

Answer 1

To be given in question

$T_{1}=27°=300K$

$Pressure P_{1}=1atm$

$Pressure P_{2} =50atm$

$\gamma=\frac{5}{3}$

To be asked in question

$T_{2}=?$

We know that

$PT^\frac{\gamma}{\gamma-1}. =constant$

$(\frac{P_{1}}{P_{2}})^{1-\gamma}=(\frac{T_{2}}{T_{1}})^{\gamma}$

Put value

$(\frac{1}{50})^{1-\frac{5}{3}}=(\frac {T_{2}}{300})^\frac{5}{3}$

$(\frac{1}{50})^{-\frac{2}{3}}=(\frac {T_{2}}{300})^\frac{5}{3}$

$({50})^{\frac{2}{3}}=(\frac {T_{2}}{300})^\frac{5}{3}$

$T_{2}=1434.52K$

$T_{2}=1161°C$