A diesel engine has a compression ratio of 20 | vedclass

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(B) The compression ratio is given by $r = V_1/V_2 = 20$. Thus,$V_2 = V_1/20 = (10^{-3}/20) \ m^3$. For an adiabatic process,$p_1 V_1^\gamma = p_2 V_2

Vedclass · Accepted Answer

$-579$

Vedclass · Answer

(B) The compression ratio is given by $r = V_1/V_2 = 20$. Thus,$V_2 = V_1/20 = (10^{-3}/20) \ m^3$. For an adiabatic process,$p_1 V_1^\gamma = p_2 V_2^\gamma$. Therefore,$p_2 = p_1 (V_1/V_2)^\gamma = 10^5 	imes (20)^{1.4} = 10^5 	imes 66.3 = 66.3 	imes 10^5 \ Pa$. The work done by the gas during an adiabatic process is given by $W = \frac{p_1 V_1 - p_2 V_2}{\gamma - 1}$. Substituting the values: $W = \frac{(10^5 	imes 10^{-3}) - (66.3 	imes 10^5 	imes 10^{-3} / 20)}{1.4 - 1}$. $W = \frac{100 - 331.5}{0.4} = \frac{-231.5}{0.4} = -578.75 \ J \approx -579 \ J$. Since the work is done on the gas,the value is negative.

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