Part A Compute the specific heat capacity at constant volume of nitrogen (N2) ga

Question

Part A

Compute the specific heat capacity at constant volume of nitrogen (N2) gas. The molar mass of N2 is 28.0 g/mol.

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Part B

You warm 1.05 kg of water at a constant volume from 23.0 C to 32.0 C in a kettle. For the same amount of heat, how many kilograms of 23.0 C air would you be able to warm to 32.0 C? Make the simplifying assumption that air is 100% N2.

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Part C

What volume would this air occupy at 23.0 C and a pressure of 1.08 atm ?

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Explanation / Answer

(A) Molar heat capacity of N2 (diatomic gas) at constant volume cv = 5/2*R = 20.7 J/mol-K

hence specific heat capacity of N2 at constant volume c = cv/28*10-3 = 20.7/28*10-3 = 739.3 J/Kg-K

(B) specific heat capacity of water C = 418 J/Kg-0C

hence heat required to heat 1.05 Kg of water from 230C to 320C (Q) = 1.05*418*(32-23) = 3950.1 J

so according to the problem Q = m*c*(32-23) where m is the mass of air to be heated

hence 3950.1 = m*739.3*9 or m = 0.594 Kg

(C) we have T= 23+273 = 296 K ; pressure P = 1.08 atm and number of moles n = 594/28 = 21.2

therefore applying gas law

PV = nRT or V = 21.2*0.082*296/1.08 = 476.45 L

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