Question

An ideal gas initially at 4.00atm and 350 K is permitted

Answer 1

Explanation:

It is given that initially pressure of ideal gas is 4.00 atm and its temperature is 350 K. Let us assume that the final pressure is $P_{2}$ and final temperature is $T_{2}$.

(a)   We know that for a monoatomic gas, value of $\gamma$ is \frac{5}{3}[/tex].

And, in case of adiabatic process,

                $PV^{\gamma}$ = constant              

also,         PV = nRT

So, here    $T_{1}$ = 350 K,    $V_{1} = V$,  and   $V_{2} = 1.5 V$

Hence,      $\frac{T_{2}}{T_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma -1}$

         $\frac{T_{2}}{350 K} = (\frac{V}{1.5V})^{\frac{5}{3} -1}$

          $T_{2}$ = 267 K

Also,   $P_{1}$ = 4.0 atm,   $V_{1} = V$,  and   $V_{2} = 1.5 V$

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

        $\frac{P_{2}}{4.0 atm} = (\frac{V}{1.5V})^{\frac{5}{3}}$

            $P_{2}$ = 2.04 atm

Hence, for monoatomic gas final pressure is 2.04 atm and final temperature is 267 K.

(b) For diatomic gas, value of $\gamma$ is \frac{7}{5}[/tex].

As,        $PV^{\gamma}$ = constant              

also,         PV = nRT

$T_{1}$ = 350 K,    $V_{1} = V$,  and   $V_{2} = 1.5 V$

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

         $\frac{T_{2}}{350 K} = (\frac{V}{1.5V})^{\frac{7}{5} -1}$

          $T_{2}$ = 289 K

And,   $P_{1}$ = 4.0 atm,   $V_{1} = V$,  and   $V_{2} = 1.5 V$

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

        $\frac{P_{2}}{4.0 atm} = (\frac{V}{1.5V})^{\frac{7}{5}}$

            $P_{2}$ = 2.27 atm

Hence, for diatomic gas final pressure is 2.27 atm and final temperature is 289 K.