the pressure of a gas change linearly with volume from 10kPa, 200 cc to 50 kPa, 50 cc. calculate the work done by the gas.

Answer

A, B, D

Explanation

`P_10 kP_a`
`=`10xx10^3P-a`
`P_2=50xx10^3P_a`
`V_1=200CC, V_2=50CC`
`(i)work done on the gasltbrge`[((1)/(2))(10+50)xx10^3xx(50-200)xx10^-6]`
`=`-`4*5 J`
`(ii)`DeltaQ=0,`
`So,lt `DeltaU=-DeltaW=4*5J`