Model air as a diatomic ideal gas with M = 28.9 g/mol. A cylinder with a piston

Question

Model air as a diatomic ideal gas with M = 28.9 g/mol. A cylinder with a piston contains 1.15 kg of air at 27.0°C and 1.55 105 Pa. Energy is transferred by heat into the system as it is permitted to expand, with the pressure rising to 3.55 105 Pa. Throughout the expansion, the relationship between pressure and volume is given by P = CV^(1/2) where C is a constant

a) Find the initial volume

b) Find the final volume

c) Find the final temperature

d) Find the work done on the air

e) Find the energy transferred by heat

Explanation / Answer

  a.
The number of moles of the gas is
n = m/M = 1150g / 28.9g/mol = 39.79mol

Initial volume can be found from ideal gas law:
PiVi = nRTi
=>
Vi = nRTi / Pi
= 39.79mol 8.31447Pam³/molK (27 + 273.15K) / 1.55*10^5 Pa
= 0.6406m³


b.
P = CV
=>
Pi /Vi = C = Pf /Vf
=>
Vf = Vi (Pf/Pi)²
= 0.6406m³ (3.55*10^5Pa / 1.55*10^5Pai)²
= 3.3603m³


c.
PfVf = nRTf
=>
Tf = PfVfM /(nR)
= 3.55*10^5Pa 3.3603m³ *28.9/(39.79mol 8.314472J/molK)
= 104212.6K
= 103939.6°C


d.
W = - P dV from Vi to Vf
= - CV dV from Vi to Vf
= - C(2/3) (Vf³ - Vi³ )
< with C = Pi /Vi
= - (2/3) (Pi /Vi) (Vf³ - Vi³ )
= - (2/3)PiVi((Vf/Vi)³ - 1
= - (2/3) 1.55*10^5Pa 0.6406m³ ((3.3603/0.6406)³ - 1 )
= - 791766Pam³
= - 792 KJ


e.
The change of internal energy of the gas equals heat transferred to plus work done on it:
U = Q + W
=>
Q = U - W

The change of internal energy of an ideal gas is
U = nCvT
For a diatomic gas
Cv = (5/2)R
Hence:
U = (5/2)nR(Tf - Ti)
= 39.79mol (5/2) 8.314472J/molK (103939.6 - 27)
= 8593kJ

=>

Q = 8593kJ - (-792kJ) = 9385kJ

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