Calculating Volume & Work: V1, V2, W

[miekier]

Homework Statement

500 kg of O2 gas with a molar mass of 32 g/mol is compressed by a polytropic compressor with constant k = 1.3. Initial pressure is 1 bar, pressure after compression is 3 bar. The initial temperature is 15°C = 288.15 K. This process takes 50 minutes. Calculate the work done and the power.

Relevant Equations

P = W/t
p*V^(k) = constant
W = (p1*V1-p2*V2)/(k-1)
p*V = n*R*T

First i calculate n:
500000 g / 32 g/mol = 15625 mol

Then I calculate V1:
(100000 Pa * V1)/288.15 K = 15625 mol * 8.31 J/mol.k
so V1 = 374.145 m^3

I calculate V2:
100000¨ Pa * (374.145 m^3)^1.3 = 300000 Pa * (V2)^1.3
So V2 = 160.703 m^3

W = ((100000 Pa * 374.145 m^3)-(300000 Pa * 160.703 m^3))/(1.3-1) = -35988000 J?

I know my work is supposed to be positive, and the only value that can be wrong is V2 I think? Although that calculation seems fine to me, so I don't understand why the result for W is negative.

 

[Steve4Physics]

W = (p1*V1-p2*V2)/(k-1)W = ((100000 Pa * 374.145 m^3)-(300000 Pa * 160.703 m^3))/(1.3-1) = -35988000 J?

Not a familiar topic but in your equation "W = (p1*V1-p2*V2)/(k-1)" should "k-1" actually be "1-k"?

Also, it is good practice to give answer such as "35988000J" in standard form (scientific notation) rounded to an appropriate number of significant figures!

 

[miekier]

Well in our textbook the formula is listed with (k-1), but I can't find any similar formula online so I can't double check.
Although in the example problem the teacher uses the same formula and finds a positive value for W. That's what makes me think that my V2 must be wrong.
You're right about the scientific notation being better, it's just laziness with typing exponents that makes me not round them immediately.
At any rate, thank you for taking a look at the problem.

 

[Steve4Physics]

Well in our textbook the formula is listed with (k-1), but I can't find any similar formula online so I can't double check.
Although in the example problem the teacher uses the same formula and finds a positive value for W. That's what makes me think that my V2 must be wrong.
You're right about the scientific notation being better, it's just laziness with typing exponents that makes me not round them immediately.
At any rate, thank you for taking a look at the problem.

Any chance there's confusion between the work done by the gas and the work done on the gas? They are of course negatives of each other.

 

[Arjan82]

I think it might be correct, and you've just computed the work done by the gas, on the system. Which is indeed negative.

Steve4Physics beat me to it :)