You warm 1.00 kg of water at a constant volume of 1.00 L from 20.0 degrees C to 30.0 degrees C in a kettle. For the same amount of heat, how many kilo grams of 20.0 degree C air would you be able to warm to 30.0 degree C? What volume would this air occupy at 20.0 degrees C and a pressure of 1.00 atm? Make the simplifying assumption that air is 100% N2.(adsbygoogle = window.adsbygoogle || []).push({});

2. Relevant equations

m = 2CvQ/(3RΔTcv)

ρ=m/v

dQ = mcdT

3. The attempt at a solution

Q = (1.00kg)(4190 J/mol*K)(303K-293K) = 41900 J

c(air) = 1.01 * 10^3 J/kg*K

V= 1.00 L

ρ(air) =1.2929 kg/m^-3

m(air) = ρ(air)*V <=>

m(air) = (1.2929 kg/m^-3)(1.00 L) = 1.2929 kg

C(air) = (1.01*10^3 J/kg*K)(1.2929 kg) = 1305.829 J/mol*K

R = 8.314 J/mol*K

cv = 4190 J/mol*K

the answer should be 5.65 kg

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# Compute Heat capacity

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