Question

One mole of an ideal gas is put through a series of changes as shown in the figure in which 1, 2, 3 mark the three stages of the system. Find the pressure (in bar) at three stages.

• A. $\begin{array}{ccc}1& 2& 3\\ 1.107& 2.214& 1.107\end{array}$
• B. $\begin{array}{ccc}1& 2& 3\\ 1.113& 2.226& 2.226\end{array}$
• C. $\begin{array}{ccc}1& 2& 3\\ 2.214& 1.107& 2.214\end{array}$
• D. $\begin{array}{ccc}1& 2& 3\\ 1.53& 3.06& 3.06\end{array}$

Answer 1

The correct option is B $\begin{array}{ccc}1& 2& 3\\ 1.113& 2.226& 2.226\end{array}$

At stage (1),
$V=22.4L=22.4\times {10}^{-3}{m}^3$
$T=300K,n=1mol$
$P_1=\frac{nRT}{V}=\frac{1mol\times 8.314Pa{m}^{3}mo{l}^{-1}{K}^{-1}\times 300K}{22.4\times {10}^{-3}{m}^3}$
$1.113\times {10}^{5}Pa=1.113bar$

At state (2), temperature is doubled but the volume is constant.
Thus, $asP\alpha T,P_2=2P_1=2.226bar$

At state (3), T is 300 K and V is 11.2 L.
So, putting the values in the formula as we did in stage (1),
$P=\frac{nRT}{V}=2.226bar$