Question

$2{\text{mm}}^3$ of a gas is compressed adiabatically at $1\text{Bar}$ pressure from temperature ${27}^{\circ }\text{C}$ to ${627}^{\circ }\text{C}$. Find the final pressure of the gas. Adiabatic index for the gas is $1.5$

• A. $27\times {10}^5{\text{N/m}}^2$
• B. $80\times {10}^5{\text{N/m}}^2$
• C. $36\times {10}^5{\text{N/m}}^2$
• D. $56\times {10}^5{\text{N/m}}^2$

Answer 1

The correct option is A $27\times {10}^5{\text{N/m}}^2$

Initial pressure $P_1=1\text{Bar}={10}^5{\text{N/m}}^2$
Initial Temperature, $\left(T_1\right)={27}^{\circ }\text{C}=300\text{K}$
Final Temperature, $\left(T_2\right)={627}^{\circ }\text{C}=900\text{K}$
Adiabatic index of gas, $\gamma =1.5$
For an adiabatic process,
$P^{1-\gamma }{T}^{\gamma }=k$
$\Rightarrow P_1^{1-\gamma }{T}_1^{\gamma }=P_2^{1-\gamma }{T}_2^{\gamma }$
$P_2=P_1{\left(\frac{T_1}{T_2}\right)}^{\frac{\gamma }{1-\gamma }}$
$={10}^5{\left(\frac{300}{900}\right)}^{-3}$
$=27\times {10}^5{\text{N/m}}^2$