Question

$1\, mm ^{3}$ of a gas is compressed at 1 atmospheric pressure and from temperature $27^{\circ} C$ to $627^{\circ} C$. What is the final pressure under adiabatic condition $(\gamma$ for the gas $=1.5)$ ?

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

Answer 1

Answer: (A)

Here, $P=10^{5} N / m ^{2}, T_{1}=27+273=300\, K$
$T_{2}=627+273=900\, K$ and $\gamma=1.5$
For adiabatic change, $\frac{T_{\gamma}}{p^{\gamma-1}}=$ constant
$\left(\frac{p_{2}}{p_{1}}\right)^{1 / 2}=\left(\frac{T_{2}}{T_{1}}\right)^{3 / 2}$
$\Rightarrow \left(\frac{p_{2}}{105}\right)^{1 / 2} =\left(\frac{900}{300}\right)^{3 / 2}$
$\Rightarrow p_{2} =27 \times 10^{5} N / m ^{2}$