Question

An ideal gas, initially at a pressure of 11.2 atm and a temperature of 299 K, is allowed to expand adiabatically until its volume doubles.

Required:
What is the gas’s final pressure, in atmospheres, if the gas is diatomic?

Answer 1

Answer:

The pressure is  $P_2 = 4.25 \ a.t.m$

Explanation:

From the question we are told that

   The initial pressure is $P_1 = 11.2\ a.t.m$

   The  temperature is  $T_1 = 299 \ K$

   

Let the first volume be  $V_1$ Then the final volume will be  $2 V_1$

 Generally for a diatomic  gas

           $P_1 V_1 ^r = P_2 V_2 ^r$

Here r is the radius of the molecules which is  mathematically represented as

    $r = \frac{C_p}{C_v}$

Where $C_p \ and\ C_v$ are the molar specific heat of a gas at constant pressure and  the molar specific heat of a gas at constant volume with values

     $C_p=7 \ and\ C_v=5$

=>   $r = \frac{7}{5}$

=>  $11.2*( V_1 ^{\frac{7}{5} } ) = P_2 * (2 V_1 ^{\frac{7}{5} } )$

=>   $P_2 = [\frac{1}{2} ]^{\frac{7}{5} } * 11.2$

=>  $P_2 = 4.25 \ a.t.m$

Answer 2

The final pressure of the gas will be 4.244 atm.

Given information:

The initial pressure of the gas is $P_1=11.2\rm\;atm$.

The initial temperature of the gas is $T_1=299\rm\; K$.

Let the initial volume of the gas be V. So, the final volume will be double or 2V.

The given diatomic gas is expanded adiabatically. So, the equation of the adiabatic process will be,

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where $\gamma$ is the ratio of specific heats of the gas which is equal to 1.4 for a diatomic gas.

So, the final pressure $P_2$ can be calculated as,

$P_1V_1^{\gamma}=P_2V_2^{\gamma}\\11.2\times V^{1.4}=P_2\times (2V)^{1.4}\\P_2=4.244\rm\;atm$

Therefore, the final pressure of the gas will be 4.244 atm.

For more details, refer to the link:

https://brainly.com/question/16014998