Question

A mass of diatomic gas $(\gamma =1.4)$ at a pressure of $2atm$ 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. $28atm$
• B. $68.7atm$
• C. $256atm$
• D. $8atm$

Answer 1

Answer: (C)

$256atm$

$P{V}^{\gamma }=$ constant $T_1=273+27=300K$
$P{\left(\frac{T}{P}\right)}^{\gamma }=$ constant $T_2=273+927=1200K$

$P^{1-\gamma }{T}^{\gamma }=$ constant
$\Rightarrow P_1^{1-\gamma }{T}_1^{\gamma }=P_2^{1-\gamma }{T}_2^{\gamma }$
$\Rightarrow 2^{1-1.4}{\left(300\right)}^{1.4}=P_2^{1-1.4}\cdot {\left(1200\right)}^{1.4}$
$\Rightarrow {\left(\frac{P_2}{P_1}\right)}^{1-\gamma }={\left(\frac{T_1}{T_2}\right)}^{\gamma }\Rightarrow \frac{P_2}{P_1}={\left(\frac{T_1}{T_2}\right)}^{\frac{\gamma }{1-\gamma }}$

${\left(\frac{P_1}{P_2}\right)}^{1-\gamma }={\left(\frac{T_2}{T_1}\right)}^{\gamma }$

${\left(\frac{P_1}{P_2}\right)}^{1-1.4}={\left(\frac{1200}{300}\right)}^{1.4}$

${\left(\frac{P_1}{P_2}\right)}^{-0.4}={\left(4\right)}^{1.4}$

${\left(\frac{P_2}{P_1}\right)}^{0.4}=4^{1.4}$

$P_2=P_1{4}^{\left(1.4/0.4\right)}=P_1{4}^{\left(7/2\right)}$

$=P_1\left(2^7\right)=2\times 128=256$