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

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

Answer 2

hello,
we know that,
mass of nitrogen=2×10⁻²kg=20g
rise in temperature=ΔT=45°C
heat required =Q=?
Q=nCT
C=7R/2(diatomic molecule)
C=7×8.3/2
n(no. of moles)=w/m
w=20g m=28u
n=20/28
n=1/1.4 moles
let the temp. be 45K
Q=10/14×7/2×8.3×45K
Q=933.75 J
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