3.6 Adiabatic processes for an ideal gas  (Page 2/5)

Compression of an ideal gas in an automobile engine

Gasoline vapor is injected into the cylinder of an automobile engine when the piston is in its expanded position. The temperature, pressure, and volume of the resulting gas-air mixture are $20\text{°}\text{C}$ , $1.00\times {10}^5{\text{N/m}}^2\text{,}$ and $240{\text{cm}}^3$ , respectively. The mixture is then compressed adiabatically to a volume of $40{\text{cm}}^3$ . Note that in the actual operation of an automobile engine, the compression is not quasi-static, although we are making that assumption here. (a) What are the pressure and temperature of the mixture after the compression? (b) How much work is done by the mixture during the compression?

Strategy

Because we are modeling the process as a quasi-static adiabatic compression of an ideal gas, we have $p{V}^{\gamma }=\text{constant}$ and $pV=nRT$ . The work needed can then be evaluated with $W={\displaystyle {\int }_{V_1}^{V_2}pdV}$ .

Solution

1. For an adiabatic compression we have
   
   $p_2=p_1{\left(\frac{V_1}{V_2}\right)}^{\gamma },$
   
   so after the compression, the pressure of the mixture is
   
   $p_2=(1.00\times {10}^5{\text{N/m}}^2){\left(\frac{240\times {10}^{\mathrm{-6}}{\text{m}}^3}{40\times {10}^{\mathrm{-6}}{\text{m}}^3}\right)}^{1.40}=1.23\times {10}^6{\text{N/m}}^2\text{.}$
   
   From the ideal gas law, the temperature of the mixture after the compression is
   
   $\begin{array}{cc}\hfill T_2& =\left(\frac{p_2{V}_2}{p_1{V}_1}\right)T_1\hfill \\ & =\frac{(1.23\times {10}^6{\text{N/m}}^2)(40\times {10}^{\mathrm{-6}}{\text{m}}^3)}{(1.00\times {10}^5{\text{N/m}}^2)(240\times {10}^{\mathrm{-6}}{\text{m}}^3)}·293\text{K}\hfill \\ & =600\text{K}=328\text{°}\text{C}\text{.}\hfill \end{array}$
2. The work done by the mixture during the compression is
   
   $W={\displaystyle {\int }_{V_1}^{V_2}pdV}.$
   
   With the adiabatic condition of [link] , we may write p as $K\text{/}{V}^{\text{γ}},$ where $K=p_1{V}_1^{\gamma }=p_2{V}_2^{\gamma }.$ The work is therefore
   
   $\begin{array}{cc}\hfill W& ={\displaystyle {\int }_{V_1}^{V_2}\frac{K}{V^{\gamma }}}dV\hfill \\ & =\frac{K}{1-\gamma }\left(\frac{1}{V_2{}^{\gamma -1}}-\frac{1}{V_1{}^{\gamma -1}}\right)\hfill \\ & =\frac{1}{1-\gamma }\left(\frac{p_2{V}_2^{\gamma }}{V_2{}^{\gamma -1}}-\frac{p_1{V}_1^{\gamma }}{V_1{}^{\gamma -1}}\right)\hfill \\ & =\frac{1}{1-\gamma }(p_2{V}_2-p_1{V}_1)\hfill \\ & =\frac{1}{1-1.40}[(1.23\times {10}^6{\text{N/m}}^2)(40\times {10}^{\mathrm{-6}}{\text{m}}^3)\hfill \\ & -(1.00\times {10}^5{\text{N/m}}^2)(240\times {10}^{\mathrm{-6}}{\text{m}}^3)]\hfill \\ & =\mathrm{-63}\text{J}\text{.}\hfill \end{array}$

Significance

The negative sign on the work done indicates that the piston does work on the gas-air mixture. The engine would not work if the gas-air mixture did work on the piston.

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