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) $\left(C_V=20.8J/molK,{\gamma }_{air}=1.4,(20{)}^{1.4}=66.3\right)$

• 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,
$W=\frac{p_1{V}_1-p_2{V}_2}{\gamma -1}$
$W=\frac{10^5\times 10^{-3}-66.3\times 10^5\times \frac{10^{-3}}{20}}{1.4-1}$
$W=-579J$