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 N 2 = 28; R = 8.3 J mol –1 K –1 .)

Question

Philoid · Accepted Answer

we know amount of heat transferred to any substance (solid , liquid, Gas) at constant pressure is given by the relation

ΔQ = nC_PΔT

Where ΔQ is the amount of heat supplied to the substance in raising it temperature by ΔT, C_P is the molar specific heat of substance at constant pressure, n is the number of moles of substance.

Here the Substance is Nitrogen Gas (N₂) , which is a diatomic gas, for a diatomic Gas the molar specific heat is

R is universal Gas Constant , R = 8.3 J mol^–1 K^–1

The rise in temperature is of 45°C, so we have

ΔT = 45°C = 45 K

(since we are only concerned with difference of temperature so unit can be either ^oC or K , both have same interpretation)

To find the moles of N₂ Gas we have the relation

n = m/M

where n is the number of moles of gas, m is the mass of gas in grams (g) and M is the Molar mass of the gas in g/mol

here we are given mass of N₂ as

m = 2.0 × 10^–2 kg

1 Kg = 1000 g , so we have

m = 2.0 × 10^–2 × 10³ = 20 g

molar mass of N₂ gas is

M = 28 g/mol

So the number of moles of N₂ gas is

so putting the value of ΔT,n,C_P in the equation ΔQ = nC_PΔT

ΔQ = 0.714mol ×(7/2) × R × 45 K

= 0.714mol × 3.5 × 8.3 J mol^–1 K^–1 × 45K

= 933.38 J

So 933.38 J heat must be supplied to 20g nitrogen gas to raise its temperature by 45^oC at constant Pressure

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.)

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.)