`14 g of N_(2)` gas is heated in a closed rigid container to increase its temperature from `23^(@)C` to `43^(@)C`. The amount of heat supplied to the gas is

Options

A. `25 cal`
B. `50 cal`
C. `100 cal`
D. `30 cal`

Answer

B

Explanation

`n = m/M=(14g)/(28g//mol) = 1/2 mol`.
In a closed rigid container `DeltaV = 0`, and hence, `DeltaW =0`
so, `DeltaQ=DeltaU = nC_vDeltaT=n((5R)/(2))DeltaT`
`=(1/2 mol)(5/2) xx (2 cal)/("mol" -^@C)) (43-23)^(@)C = 50cal`.