Question

A mass of diatomic gas ($\gamma =1.4$) at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rise from 27${}^o$C to 927${}^o$C. The pressure of the gas is final state is

• A. 28 atm
• B. 68.7atm
• C. 256 atm
• D. 8 atm

Answer 1

The correct option is C 256 atm

For an adiabatic process
$\frac{T^{\gamma }}{P^{\gamma -1}}=constant$
Given,
$T_1={27}^{o}C=273+27={300}^{o}K$
$T_2={927}^{o}C=273+927={1200}^{o}K$
$P_1=2atm$
$\gamma =1.4$

$\frac{T_1^{\gamma }}{P_1^{\gamma -1}}=\frac{T_2^{\gamma }}{P_2^{\gamma -1}}$

Substituting the values,

$\frac{{300}^{1.4}}{2^{1.4-1}}=\frac{{1200}^{1.4}}{P_2^{1.4-1}}$

$\therefore (\frac{P_2}{2}{)}^{0.4}=(\frac{1200}{300}{)}^{1.4}$

$\ln \left(\right(\frac{P_2}{2}{)}^{0.4}=\ln (\frac{1200}{300}{)}^{1.4}$

$\therefore 0.4\ln \left(\frac{P_2}{2}\right)=1.4\ln \left(\frac{4}{1}\right)={1.4}^2$

$\therefore \ln \left(\frac{P_2}{2}\right)=4.9$

$\therefore P_2=256atm$