Question

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).

Answer 1

m = 2 × 10³ kg^-1 = 20g,

∆T = 45°C, M = 28

R = 8.3 J mol^-1 K^-1, M = 28

Number of moles, n = \(\frac{m}{M} = \frac{20}{5} = \frac{5}{7}\)

Since nitrogen is diatomic,

∴ C_P = \(\frac{7}{2}\)R = \(\frac{7}{2}\) × 8.3 J mol^-1K^-1

Now, ∆Q = nC_P ∆T

\(\frac{5}{2} \times \frac{7}{2}\)× 8.3 × 45J = 933.75J.