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 1.02 kg of air at 28.0°C and 165 kPa. Energy is transferred into the system as it is allowed to expand, with the pressure rising to 365 kPa. Throughout the expansion, the relationship between pressure and volume is given by the following expression, where C is a constant.

P=CV(1/2)

(a) Find the initial volume.

m3

(b) Find the final volume.
m3

(c) Find the final temperature.
K

(d) Find the work done on the air.
J

(e) Find the energy transferred by heat.
MJ

Explanation / Answer

a:- The number of moles of the gas is
n = m/M = 1020g / 28.9g/mol = 35.29 mol
Initial volume can be found from ideal gas law:
PiVi = nRTi
=> Vi = nRTi / Pi = 35.29mol 8.31 (28 + 273.15K) / 165000 = 0.5352 m^3

b:- P = CV
=> Pi /Vi = C = Pf /Vf
=> Vf = Vi (Pf/Pi)^2 = 0.5352 m^3 (365kPa / 165kPa)^2 = 2.619 m^3

c:- PfVf = nRTf
=> Tf = PfVf/(nR) = 365000 2.619/(35.29 8.31) = 3259.69 K

d:- W = - P dV from Vi to Vf
= - CV dV from Vi to Vf
= - C(2/3) (Vf³ - Vi³ )
Since, C = Pi /Vi
So, W = - (2/3) (Pi /Vi) (Vf³ - Vi³ )
= - (2/3)PiVi((Vf/Vi)³ - 1
= - (2/3) 16500Pa 0.5352 ((2.619/0.5352)^3 - 1 )
= - 57841.91 J

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