Physics: 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?

An ideal gas, initially at a pressure of 11.2 atm

jessikamacadlo8948

31.01.2022

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

$C_p=7 \ and\ C_v=5$

=> $r = \frac{7}{5}$

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

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

=> $P_2 = 4.25 \ a.t.m$

0,0(0 оценок)