What amount of heat must be supplied to 2.0 x 10^{-2} kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N_{2} = 28; R = 8.3 J mol^{-1} K^{-1}.)

#### Solution 1

Mass of nitrogen, *m* = 2.0 × 10^{–2} kg = 20 g

Rise in temperature, Δ*T* = 45°C

Molecular mass of N_{2}, *M* = 28

Universal gas constant, *R* = 8.3 J mol^{–1} K^{–1}

Number of moles, `n = m/M`

`= (2.0xx10^(-2)xx10^3)/28 = 0.714`

Molar specific heat at constant pressure for nitrogen, `C_P = 7/2 R`

`= 7/2 xx 8.3`

`= 29.05 J mol^(-1) K^(-1)`

The total amount of heat to be supplied is given by the relation:

ΔQ = *nC*_{P }Δ*T*

= 0.714 × 29.05 × 45

= 933.38 J

Therefore, the amount of heat to be supplied is 933.38 J.

#### Solution 2

Here, mass of gas,` m= 2xx 10^-2 kg` = 20 g

rise in temperature, `triangleT = 45 ^@C`

Heat required, triangleQ = ?, Molecular mass, M = 28

Number of moles , `n = m/M = 20/28 = 0.714`

As nitrogen is a diatomatic gas,molar specific heat at constant pressure is

`C_p = 7/2R = 7/2 xx 8.3 J mol^(-1) K^(-1)`

As `triangleQ = nC_p triangleT`

`:. triangleQ = 0.714 xx 7/2 xx 8.3 xx 45 J = 933.4 J`