Question

Question: The \(pv\)-diagram in Fig. E20.5 shows a cycle of a heat engine that uses \(0.250\;mol\) of an ideal gas with \(\gamma = 1.40\). Process ab is adiabatic. (a) Find the pressure of the gas at point \(a\). (b) How much heat enters this gas per cycle, and where does it happen? (c) How much heat leaves this gas in a cycle, and where does it occur? (d) How much work does this engine do in a cycle? (e) What is the thermal efficiency of the engine?

Short answer

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

Step-by-step solution

Write the given data from the question.

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 \(c\),\({p_c} = 1.5\;{\rm{atm}}\)

Determine the formulas to calculate the pressure at the point \(a\).

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 \(a\).

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 \(12.3\;{\rm{atm}}\).