A diesel engine has a compression ratio of `20:1.` If the initial pressure is `1xx 10 ^(5)` Pa and the initial volume of the cylinder is `1 xx 10^(-3)m ^(2),` then how much work does the gas do during the compression ?
(Assume the process as adiabatic)
`(C_(v) =20.8J//mol K, Y _(ari)=1.4, (20) ^(1.4)=66.3)`

Options

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

Answer

B

Explanation

Compression ratio `= ("initial volume ")/("Final volume" ) = (20)/(1)`
`(V_(1))/(V_(2)) = (20)/(I)`
`implies V_(2) = (V_(1))/(20) = (10 ^(-3))/(20)m ^(3)`
In adiabatic process,
`p _(1) v _(1) ^(gamma) = p_(2) v _(2) ^(gamma)`
`implies p _(2) = ((V_(1))/(V_(2)))^(gamma )p _(1)`
Given, initial pressure, `p _(1) = 10^(5) Pa, gamma = 1.4`
`implies p _(2) = ((20)/( 1))^(1.4) xx 10 ^(5)=66.3 xx 10^(5) Pa`
`implies ` Now, work done in adiabatic compression by gas,
`w = (p_(1) V_(1) - p_(2) V_(2))/(gamma -1)`
`W = (10^(5) xx 10 ^(-3) - 66.3 xx 10 ^(5) xx (10 ^(-3))/(20))/(1.4-1)`
`W =-579J`