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