Question

A gas having adiabatic constant $\left(\gamma \right)=1.4$ and volume $2{\text{m}}^3$, is at a pressure of $4\times {10}^5{\text{N/m}}^2$. It is then compressed adiabatically to a new volume $0.5{\text{m}}^3$. Find the new pressure and the work done during the process.
[Take $4^{1.4}=6.96$]

• A. $3.47\times {10}^5{\text{N/m}}^2,-11.78\times {10}^5\text{J}$
• B. $3.47\times {10}^5{\text{N/m}}^2,-1475\times {10}^5\text{J}$
• C. $2.78\times {10}^6{\text{N/m}}^2,-11.78\times {10}^5\text{J}$
• D. $2.78\times {10}^6{\text{N/m}}^2,-14.75\times {10}^5\text{J}$

Answer 1

The correct option is D $2.78\times {10}^6{\text{N/m}}^2,-14.75\times {10}^5\text{J}$

For an adiabatic process, equation of state in terms of $P$ and $V$ is given by
$P{V}^{\gamma }=k$
From this, we can write that
$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$
or $P_2=P_1{\left(\frac{V_1}{V_2}\right)}^{\gamma }$
Given,
$P_1=4\times {10}^5{\text{N/m}}^2,V_1=2{\text{m}}^3,V_2=0.5{\text{m}}^3,P_2=?$
Thus,
$P_2=(4\times {10}^5)\times {\left(\frac{2}{0.5}\right)}^{1.4}$
$\Rightarrow P_2=2.78\times {10}^6{\text{N/m}}^2$
Now work done during an adiabatic process is given by,
$W=\frac{1}{1-\gamma }(P_2{V}_2-P_1{V}_1)$
Substituting the values, we get
$W=\frac{1}{1-1.4}\left[\right(27.80\times {10}^5)\times 0.5-(4\times {10}^5)\times 2]$
$W=-14.75\times {10}^5\text{J}$
Thus, option (d) is the correct answer.