91Ó°ÊÓ

The number of the moles,\(n = 0.250\;{\rm{g}}\)

Ratio of heat capacity,\(\gamma = 1.4\)

The pressure at point \(b\),\({p_b} = 1.5\;{\rm{atm}}\)

Volume at point \(b\),\({V_b} = 0.0090\;{{\rm{m}}^3}\)

Volume at point \(a\),\({V_a} = 0.0020\;{{\rm{m}}^3}\)

Volume at point \(c\),\({V_b} = 0.0020\;{{\rm{m}}^3}\)

The pressure at point \(c\),\({p_c} = 1.5\;{\rm{atm}}\)

The product of the pressure and volume is constant.\(\left( {pV = {\rm{constant}}} \right)\).

The expression to calculate the pressure at the point \(a\) is given as follows.

\(\begin{array}{c}{p_a} = \frac{{{p_b}V_b^\gamma }}{{V_a^\gamma }}\\{p_a} = {p_b}{\left( {\frac{{{V_b}}}{{{V_a}}}} \right)^\gamma }\end{array}\) …… (i)

Here,\({p_b}\)is the pressure at the point\(b\),\({V_b}\)is the volume at point\(b\)and\({V_a}\)is the volume at point\(a\).

Calculate the pressure at the point \(a\).

Substitute \(0.0090\;{{\rm{m}}^3}\) for \({V_b}\),\(0.0020\;{{\rm{m}}^3}\) for \({V_a}\) and \(1.5\;{\rm{atm}}\) for \({p_b}\) into equation (i).

\(\begin{array}{l}{p_a} = \left( {1.5\;{\rm{atm}}} \right){\left( {\frac{{0.0090\;{{\rm{m}}^3}}}{{0.0020\;{{\rm{m}}^3}}}} \right)^{1.4}}\\{p_a} = \left( {1.5\;{\rm{atm}}} \right){\left( {\frac{{90}}{{20}}} \right)^{1.4}}\\{p_a} = \left( {1.5\;{\rm{atm}}} \right){\left( {4.5} \right)^{1.4}}\\{p_a} = 12.3\;{\rm{atm}}\end{array}\)

Hence the pressure at the point \(a\) is \(12.3\;{\rm{atm}}\).