Question

A diesel engine has a compression ratio of $20: 1 .$ If the initial pressure is $1 \times 10^{5}\, Pa$ and the initial volume of the cylinder is $1 \times 10^{-3} m ^{3}$, then how much work does the gas do during the compression? (Assume the process as adiabatic) $\left(C_{V}=20.8\, J\, / mol\, K , \gamma_{ air }=1.4,(20)^{1.4}=66.3\right)$

• A. -880 J
• B. -579 J
• C. 220 J
• D. 485 J

Answer 1

Answer: (B)

Compression ratio
$=\frac{\text { Initial volume }}{\text { Final volume }}=\frac{20}{1}$
$\frac{V_{1}}{V_{2}}=\frac{20}{1}$
$\Rightarrow V_{2}=\frac{V_{1}}{20}=\frac{10^{-3}}{20} m ^{3}$
In adiabatic process,
$p_{1} V_{1}^{\gamma}=p_{2} V_{2}^{\gamma}$
$\Rightarrow p_{2} =\left(\frac{V_{1}}{V_{2}}\right)^{\gamma} p_{1}$
Given, initial pressure, $p_{1}=10^{5} Pa , \gamma=1.4$
$\Rightarrow p_{2}=\left(\frac{20}{1}\right)^{1.4} \times 10^{5}$
$=66.3 \times 10^{5}\, Pa$
$\Rightarrow $ Now, work done in adiabatic compression by gas,
$W=\frac{p_{1} V_{1}-p_{2} V_{2}}{\gamma-1}$
$W=\frac{10^{5} \times 10^{-3}-66.3 \times 10^{5} \times \frac{10^{-3}}{20}}{1.4-1}$
$W=-579\, J$