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A diesel engine has a compression ratio of $20: 1$. If the initial pressure is $1 \times 10^5 \mathrm{~Pa}$ and the initial volume of the cylinder is $1 \times 10^{-3} \mathrm{~m}^3$, then how much work does the gas do during the compression?
(Assume the process as adiabatic)
$$
\left(C_V=20.8 \mathrm{~J} / \mathrm{mol} \mathrm{K}, \psi_{\text {air }}=1.4,(20)^{1.4}=66.3\right)
$$
Options:
(Assume the process as adiabatic)
$$
\left(C_V=20.8 \mathrm{~J} / \mathrm{mol} \mathrm{K}, \psi_{\text {air }}=1.4,(20)^{1.4}=66.3\right)
$$
Solution:
1903 Upvotes
Verified Answer
The correct answer is:
$-579 \mathrm{~J}$
Compression ratio $=\frac{\text { Initial volume }}{\text { Final volume }}=\frac{20}{1}$
$$
\begin{aligned}
\frac{V_1}{V_2} & =\frac{20}{1} \\
V_2 & =\frac{V_1}{20}=\frac{10^{-3}}{20} \mathrm{~m}^3
\end{aligned}
$$
In adiabatic process,
$$
\begin{aligned}
p_1 V_1^\gamma & =p_2 V_2^\gamma \\
\Rightarrow \quad p_2 & =\left(\frac{V_1}{V_2}\right)^\gamma p_1
\end{aligned}
$$
Given, initial pressure, $p_1=10^5 \mathrm{~Pa}, \gamma=1.4$
$$
\Rightarrow \quad p_2=\left(\frac{20}{1}\right)^{1.4} \times 10^5=66.3 \times 10^5 \mathrm{~Pa}
$$
$\Rightarrow$ Now, work done in adiabatic compression by gas,
$$
W=\frac{p_1 V_1-p_2 V_2}{\gamma-1}
$$
$$
W=\frac{10^5 \times 10^{-3}-66.3 \times 10^5 \times \frac{10^{-3}}{20}}{1.4-1}
$$
$$
\begin{aligned}
\frac{V_1}{V_2} & =\frac{20}{1} \\
V_2 & =\frac{V_1}{20}=\frac{10^{-3}}{20} \mathrm{~m}^3
\end{aligned}
$$
In adiabatic process,
$$
\begin{aligned}
p_1 V_1^\gamma & =p_2 V_2^\gamma \\
\Rightarrow \quad p_2 & =\left(\frac{V_1}{V_2}\right)^\gamma p_1
\end{aligned}
$$
Given, initial pressure, $p_1=10^5 \mathrm{~Pa}, \gamma=1.4$
$$
\Rightarrow \quad p_2=\left(\frac{20}{1}\right)^{1.4} \times 10^5=66.3 \times 10^5 \mathrm{~Pa}
$$
$\Rightarrow$ Now, work done in adiabatic compression by gas,
$$
W=\frac{p_1 V_1-p_2 V_2}{\gamma-1}
$$
$$
W=\frac{10^5 \times 10^{-3}-66.3 \times 10^5 \times \frac{10^{-3}}{20}}{1.4-1}
$$
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