An office room contains about 2000 moles of air . The change in the internal energy of this much air when it is cooled from `34 ^@ C ` to `24^@C` at constant pressure of 1.0 atm is
[ Use ` gamm _(" air") = 1.4 ` and universal gas constant `=8.314 J // mol -K]`

Options

A. `-19xx10 ^5 J`
B. `+19xx10^5 J`
C. `-42xx10^5J`
D. `+0.7xx10^5 J`

Answer

C

Explanation

Given ,
Number of moles of air in room (n) `= 200 =2xx10 ^3`
Temperature difference `(dT)= 24-34=- 10 ^@ C `
We know that ,
` DQ= nC_(v) dT = 2xx 10 ^3 xx (R ) /( 0.4 ) -[-10]`
` =( -2 xx 10 ^(3)xx 8.314 xx10 ) /( 0.4)`
` = ( 2xx 8.314 xx 10 ^5)/(4) =( 16.628 )/( 4 ) xx10 ^5`
`=- 4.2 xx 10 ^5 J`