Two moles of an ideal monoatomic gas at NTP are compressed adiabatically and reversibly to occupy a volume of `4.48dm^(3)`. Calculate the amount of work done, `DeltaU`, final temperature and pressure of the gas. `C_(V)` for ideal gas `=12.45` J `K^(-1)mol^(-1)`

Explanation

for an ideal gas, `gamma=(C_(P))/(C_(V))=1.667`
Initial volume, `V_(1)=2xx22.4=44.8dm^(3)`
Initial pressure, `P_(1)=1atm`
Initial temperature `T_(1)=273K`
Final volume `V_(2)=4.48dm^(3)`
Let the final pressure be `P_(2)` and temperature be `T_(2)`. ltBrgt Applying `P_(1)V_(1)^(gamma)=P_(2)V_(2)^(gamma)`
or `(P_(1))/(P_(2))=((V_(2))/(V_(1)))^(gamma)=((4.48)/(44.8))^(1.667)`
or `(P_(2))/(P_(1))=(10)^(1.667)`
`P_(2)=(10)^(1.667)(P_(1)=1" given")`
`logP_(2)=1.667log10=1.667`
`P_(2)=antilog1.667=46.45atm`
Final temperature `=(P_(2)V_(2))/(P_(1)V_(1))*T_(1)=(46.45xx4.48)/(1xx44.8)xx273`
`=1268K`
Work done on the system `=n*C_(V)*DeltaT`
`=2xx12.45xx(1268-273)`
`=2xx12.45xx995=24775.5J`
From the first law of thermodynamics,
`DeltaU=q+w=0+24775.5=24775.5J`