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 N2 = 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, ΔT = 45°C
Molecular mass of N2, 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 = nCP ΔT
= 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`
