How much heat energy should be added to a mixture of `10 g` of hydrogen and `40 g` of helium to change the temp. by `50^(@)C` kept in a closed vessel? Given `R= 2 cal//mol e K`

Options

A. `2000 cal`
B. `2500 cal`
C. `2750 cal`
D. `3000 cal`

Answer

C

Explanation

From the mixture of `H_(2) and He`,
`C_(v)=(nC_(v_(1)+nC_(v_(2))))/(n_(1)+n_(2))`
`(10/2(5/2R)+(40/4)(3/2R))/(10/2+40/4)=(27.5R)/(15)`
From `Q=nC_(v)(DeltaT)`
`Q=(5+10)((27.5R)/(15))xx50`
`=27.5xx2xx50cal.= 2759 cal`