Q20-36P A heat engine takes 0.350 mol of... [FREE SOLUTION]

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University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 3: Q20-36P (page 678)

A heat engine takes 0.350 mol of a diatomic ideal gas around the cycle shown in the pV-diagram of Fig. P20.36. Process \({\bf{1}} \to {\bf{2}}\) is at constant volume, process \({\bf{2}} \to {\bf{3}}\) is adiabatic, and process \({\bf{3}} \to {\bf{1}}\) is at a constant pressure of 1.00 atm. The value of \(\gamma \) for this gas is 1.40. (a) Find the pressure and volume at points 1, 2, and 3. (b) Calculate Q, W, and 鈭哢 for each of the three processes. (c) Find the net work done by the gas in the cycle. (d) Find the net heat flow into the engine in one cycle. (e) What is the thermal efficiency of the engine? How does this compare to the efficiency of a Carnot-cycle engine operating between the same minimum and maximum temperatures T1 and T2?

Short Answer

Expert verified

The pressure and volume at point 1, 2 and 3 are \(1\;{\rm{atm}}\)and\(8.61 \times {10^{ - 3}}\;{{\rm{m}}^3}\), \(2\;{\rm{atm}}\) and \(8.61 \times {10^{ - 3}}\;{{\rm{m}}^3}\), \(1\;{\rm{atm}}\)and\(14.1 \times {10^{ - 3}}\;{{\rm{m}}^3}\).

Step by step solution

Identification of given data

The number of moles of diatomic ideal gas is\(n = 0.350\;{\rm{mol}}\)

The pressure of process 3 to 1 is\({P_1} = 1\;{\rm{atm}}\)

The temperature at point 1 is\({T_1} = 300\;{\rm{K}}\)

The temperature at point 2 is\({T_2} = 600\;{\rm{K}}\)

The temperature at point 3 is \({T_3} = 492\;{\rm{K}}\)

Conceptual Explanation

The pressure at point 1 will be equal to the atmospheric pressure and temperature at point 1 is found by the ideal gas equation. The volume at point 2 is equal to the volume at point 1 because process from 1 to 2 is constant volume process and pressure at point 2 is found by the ideal gas equation. The pressure and volume at point 3 is found by the gas equation for adiabatic process.

Determination of pressure and volume at point 1, 2 and 3

The volume at the point 1 is given as:

\({P_1}{V_1} = nR{T_1}\)

Here,\(R\)is the universal gas constant and its value is\(8.31\;{\rm{J}}/{\rm{mol}} \cdot {\rm{K}}\).

Substitute all the values in the above equation.

\(\begin{aligned}\left( {1\;{\rm{atm}}} \right)\left( {\frac{{1.013 \times {{10}^5}\;{\rm{Pa}}}}{{1\;{\rm{atm}}}}} \right){V_1} = \left( {0.350\;{\rm{mol}}} \right)\left( {8.31\;{\rm{J}}/{\rm{mol}} \cdot {\rm{K}}} \right)\left( {300\;{\rm{K}}} \right)\\{V_1} = 8.61 \times {10^{ - 3}}\;{{\rm{m}}^3}\end{aligned}\)

The process 1 to 2 is constant volume process so the volume at point 2 is also\({V_2} = 8.61 \times {10^{ - 3}}\;{{\rm{m}}^3}\).

The pressure at the point 2 is given as:

\({P_2}{V_2} = nR{T_2}\)

Substitute all the values in the above equation.

\(\begin{aligned}{P_2}\left( {8.61 \times {{10}^{ - 3}}\;{{\rm{m}}^3}} \right) = \left( {0.350\;{\rm{mol}}} \right)\left( {8.31\;{\rm{J}}/{\rm{mol}} \cdot {\rm{K}}} \right)\left( {600\;{\rm{K}}} \right)\\{P_2} = \left( {2.03 \times {{10}^5}\;{\rm{Pa}}} \right)\left( {\frac{{1\;{\rm{atm}}}}{{1.013 \times {{10}^5}\;{\rm{Pa}}}}} \right)\\{P_2} = 2\;{\rm{atm}}\end{aligned}\)

The pressure at point 3 is also\({P_3} = 1\;{\rm{atm}}\)because process 1 to 3 is constant pressure process.

The volume at the point 3 is given as:

\({P_3}{V_3} = nR{T_3}\)

Substitute all the values in the above equation.

\(\begin{aligned}\left( {1\;{\rm{atm}}} \right)\left( {\frac{{1.013 \times {{10}^5}\;{\rm{Pa}}}}{{1\;{\rm{atm}}}}} \right){V_3} = \left( {0.350\;{\rm{mol}}} \right)\left( {8.31\;{\rm{J}}/{\rm{mol}} \cdot {\rm{K}}} \right)\left( {492\;{\rm{K}}} \right)\\{V_3} = 14.1 \times {10^{ - 3}}\;{{\rm{m}}^3}\end{aligned}\)

Therefore, the pressure and volume at point 1, 2 and 3 are \(1\;{\rm{atm}}\)and\(8.61 \times {10^{ - 3}}\;{{\rm{m}}^3}\), \(2\;{\rm{atm}}\) and \(8.61 \times {10^{ - 3}}\;{{\rm{m}}^3}\), \(1\;{\rm{atm}}\)and\(14.1 \times {10^{ - 3}}\;{{\rm{m}}^3}\).