Find the Mass of the Air, and the Work and Heat Transfer

[Northbysouth]

Homework Statement

A piston-cylinder assembly contains air, initially at 2 bar, 300 K, and a volume of 2m³. The air undergoes a process to a state where it pressure is 1 bar, during which the pressure-volume relationship is pv = constant. Assuming idea gas behavior for the air, determine the mass of the air, in kg and the work and heat transfer, each in kJ

Homework Equations

pv-nrt

W = ∫p dv

The Attempt at a Solution

I found the constant C to be:

p_iv_i = (200,000 Pa)(2m³)

C = 400,000 Pam³


v_f = C/p_f

Hence v_f = 400,000 Pa m³/100,000 Pa

v_f = 4m³

W = ∫P dv = ∫₂⁴C/v dv

W = 277258.9 kJ

Mass of air:

pv =nrt

n = (200,000 Pa)(2m³)/(8.314 J/K mol)(300K)

n = 160.372 Pa m³mol/J

I'm a little fuzzy on moles and such, but unless I'm much mistaken, I should be able to find the mass of the air by dividing by Avogadro number, right?

After this, I can calculate the heat transfer with Q = mcΔT in which I can find the ΔT with the relationship:

p_iv_i/T_i = p_fv_f/T_f

Am I making sense?

 

[Andrew Mason]

Homework Equations

pv=nrt

W = ∫p dv

The Attempt at a Solution

I found the constant C to be:

p_iv_i = (200,000 Pa)(2m³)

C = 400,000 Pam³

Units of Pressure x volume = Joules

v_f = C/p_f

Hence v_f = 400,000 Pa m³/100,000 Pa

v_f = 4m³

W = ∫P dv = ∫₂⁴C/v dv

Since PV = nRT and PV = constant what kind of process is this (hint: what happens to T during this process?)?

W = 277258.9 kJ

Mass of air:

pv =nrt

n = (200,000 Pa)(2m³)/(8.314 J/K mol)(300K)

n = 160.372 Pa m³mol/J

I'm a little fuzzy on moles and such, but unless I'm much mistaken, I should be able to find the mass of the air by dividing by Avogadro number, right?

Use the mass of one mole of dry air and multiply by the number of moles you have found (which is correct).

You can look up the molar mass of dry air. Or you can work it out. Dry air consists of mostly nitrogen (78%) and oxygen (21%). There is about 1% argon and traces of other gases. You have to know that the Nitrogen and oxygen gases are diatomic and argon is monatomic. 14x2(.78)+ 16x2(.21) + 40x1(.01) = 29 g/mol

After this, I can calculate the heat transfer with Q = mcΔT in which I can find the ΔT with the relationship:

p_iv_i/T_i = p_fv_f/T_f

Am I making sense?

You don't need to work out the temperature change. It is easily seen what happens to T from PV = nRT = constant.

What you need to do here is apply the first law:Q = ΔU + W

Given ΔT what can you say about ΔU? So what is Q?

AM