Question

An ideal gas at ${77}^{o}C$ is compressed adiabatically and reversibly to $\frac{9}{16}$ of its original volume. If $\gamma =\frac{5}{3},$ then the rise in temperature is:
Given: ${1.33}^{1.33}=1.46$

• A. $450K$
• B. $375K$
• C. $313K$
• D. $161K$

Answer 1

The correct option is D $161K$

For an adiabatic process
$\frac{T_2}{T_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$
where, $T_1=$ initial temperature
$T_2=$ final temperature
$V_1=$ initial volume
$V_2=$ final volume $=\frac{9}{16}{V}_1$
$\gamma =\frac{5}{3}$
$\Rightarrow T_2=350{\left(\frac{16}{9}\right)}^{\frac{5}{3}-1}=350{\left(\frac{16}{9}\right)}^{\frac{2}{3}}$
$=350{\left(\frac{4}{3}\right)}^{\frac{4}{3}}=511K$
therefore rise in tempearture is :
$\Rightarrow △T=513-350=161K$