Question

18.43---- a) You warm 2.00 kg of water at a constant volume from 19.5 ∘C to 31.0 ∘C in a kettle. For the same amount of heat, how many kilograms of 19.5 ∘C air would you be able to warm to 31.0 ∘C? Make the simplifying assumption that air is 100% N2.

b)What volume would this air occupy at 19.5 ∘C and a pressure of 1.20 atm ?

Answer 1

Part A.

Amount of heat required to heat water will be given by:

Q = Mw*Cw*dT

Cw = Specific heat capacity of water = 4186 J/kg-C

Mw = mass of water = 2.00 kg

dT = 31.0 - 19.5 = 11.5 C

So,

Q = 2.00*4186*11.5 = 96278 J

Now with this same amount of heat suppose m mass of air can be heated, then

Q = m*Ca*dT

m = mass of air = Q/(Ca*dT)

Ca = Specific heat capacity of air (Assuming 100% N2) = Cv/Mw

Cv = heat capacity at constant volume for diatomic gas = 5R/2 = 20.785 J/mol-C

Mw = Molecular weight of nitrogen = 28*10^-3 kg/mol, So

Ca = 20.785/(28*10^-3) = 742.3 J/kg-C

dT = 31.0 - 19.5 = 11.5 C

So,

m = 96278/(742.3*11.5) = 11.278 kg

m = 11.3 kg = mass of air which can be heated

Part B.

Using Ideal gas law:

PV = nRT

V = nRT/P

T = 19.5 C = 292.5 K

P = 1.20 atm = 1.20*1.01325*10^5 Pa

R = gas constant = 8.314

n = number of moles of air = m/Mw = 11.278/(28.0*10^-3) = 402.786 moles

So,

V = 402.786*8.314*292.5/(1.20*1.01325*10^5)

V = 8.06 m^3 = Volume occupied by air

Let me know if you've any query.