Compute the specific heat capacity at constant volume of nitrogen (n2) gas. the molar mass of n2 is 28.0 g/mol. c = j/(kg⋅k) request answer part b you warm 1.30 kg of water at a constant volume from 22.0 ∘c to 28.5 ∘c in a kettle. for the same amount of heat, how many kilograms of 22.0 ∘c air would you be able to warm to 28.5 ∘c? make the simplifying assumption that air is 100% n2. m = kg request answer part c what volume would this air occupy at 22.0 ∘c and a pressure of 1.10 atm ?

Part a)
Cv = M c (molar heat capacity)
where c is called specific heat and M is called Molecular weight or molar mass
c = Cv / M where Cv for air = 20.76 J/ mol.k
c = 20.76 / 28.0 x 10⁻³ = 741.43 J/ kg.k

Part b)
For the same amount of heat:
Q water = Q Nitrogen
(m.c.Δt)water = (m.c.Δt) nitrogen (Δt cancelled for the same range)
so m Nitrogen = m water x c water / c nitrogen
where Cw = 4190 and C nitrogen is 741.43 from part a)
m nitrogen = (1.30 kg * 4190) / 741.43 = 7.35 kg

Part c)
To find the volume we use:
PV = nRT
where n = mass /molar mass = 7.35 x 10³ g / 28 = 262.4 moles
R = 0.08205 
T = 22 + 273 = 295 K
P = 1.1 atm
V = nRT / P = (262.4 x 0.08205 x 295) / 1.1 = 5774 L
 

I’m not sure what your trying to say here.. can you clarify again?

The air is going to come out because of the low pressure.