Question

$1m^3$ of gas is compressed suddenly at atmospheric pressure and temperature ${27}^{o}C$ such that its temperature becomes ${627}^{o}C$. the final pressure of the gas (if $\gamma =1.5$) will be

• A. $2.7\times {10}^{5}N{m}^{-2}$
• B. $7.2\times {10}^{5}N{m}^{-2}$
• C. $27\times {10}^{5}N{m}^{-2}$
• D. $2.7\times {10}^{6}N{m}^{-2}$

Answer 1

Answer: (C)

$27\times {10}^{5}N{m}^{-2}$

For an adiabatic process, the equation of state is given by:
$T^{\gamma }{P}^{1-\gamma }=constant$
This implies,
$T_1^{\gamma }{P}_1^{1-\gamma }=T_2^{\gamma }{P}_2^{1-\gamma }$

Given:

• $T_1={27}^{\circ }C=300K$
• $T_2={627}^{\circ }C=900K$
• $P_1=1.013\times {10}^{5}Pa$
• $\gamma =1.5$

Substituting this into the equation of state:
We get $P_2=27\times {10}^{5}Pa$