Question

What amount of heat must be supplied to 2.0 × 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

$\huge \mathfrak{answer}$

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 × 10-2 × 103) / 28

= 0.714

Molar specific heat at constant pressure for nitrogen, Cp = (7/2)R

= (7/2) × 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.

Answer 2

$\huge\star\underline\mathfrak\red{Answer}$

Here,

Mass of nitrogen ( m) = 2 × 10^-2 Kg

Increase in temperature (∆T) = 45°C

molecular mass of N2 (M) = 28 g

= 28 × 10^-3 Kg

R = 8.3 J/mol.K

Number of moles (n) = m/M = 2 × 10^-2/28 × 10^-3 = 5/7

molar specific heat at constant pressure ( Cp) = YR/(Y-1) where Y is atomicity of gases . Y = 1.4

Cp = 1.4R/(1.4-1) = 7R/2

So, heat supplied = nCp∆T

= 5/7 × 7/2 × 45 × 8.3

= 933.75 J

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