Liquifying Nitrogen: How Much Heat Energy?

[says]

Homework Statement

A rigid cylinder contains 9.0g of nitrogen at 20 ∘C.

What is the minimum amount of heat energy that must be removed to liquify the nitrogen?

Homework Equations

Heat Energy
H = Cp * m * ΔT

H = Heat Energy
Cp = Heat Capacity
m = mass (in Kg)
ΔT = change in temperature (in Celsius)

Heat of Condensation (Equal to Vaporisation)

H = Hc * m

Hc= Heat of condensation
m= mass (in moles)

The Attempt at a Solution

Heat Capacity of nitrogen = 1.04 kJ/kg*K.
Nitrogen is a liquid at -196 C, so ΔT = -196 - 20 = -216

H = Heat Capacity * mass * ΔT

H = 1.04 kJ/kg*K * 0.009 kg * 215 = 2.02 kJ

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H= Heat of condensation * mass (in moles)

H= 5.56 kJ*mol * 0.32 moles = 1.78 kJ

2.02 kJ + 1.78 kJ = 3.8 kJ

3.80 kJ of heat energy is required to liquify the nitrogen.

Answer isn't correct. If anyone can point me in the right direction it would be much appreciated.

 

[BvU]

Hello says, and welcome to PF.

You seem pretty sure the answer is not correct (how do you know?), so I will accept that.

Why do you use cp if the exercise says the cylinder is rigid ?

 

[says]

Hi BvU! Thanks for the welcome. :)

I ended up getting this problem right. I was using the Cp value of Nitrogen and not the Cv value.

Problem solved!