Thermodynamics qns help!

[BOOMOOMOO]

Hi guys,

 

The question is:

Initially a piston cylinder contains air at p1=3.2bar (absolute) and the contained volume is v1=158m3 at temperatuer t1=303k. The air is then compressed, with compression ratio r=v1/v2=5.3. the polytropic index of compression is n=1.3 estimate the net heat addition in KJ.

With p2 found in my working, do i need to add my p2+p1 ?

 

Thanks

[studiot]

If you have genuinely calculated p₂ from the polytropic law then that is the compressed pressure. Pressures are not additive, unless they are partial pressures for different gases.

 

But why not tell us your calculation?

 

That is the normal etiquette for homework help.

[BOOMOOMOO]

thanks for the reply!

 

v2=158*10^-6 / 5.3 = 29.81 * 10^-6 m3

 

Using Pv=mrt to find mass
mass= 0.00058141kg

 

T2= T1(p2/p1)^0.3/1.3 = 499.7k

 

Delta U = 0.00058141 * ( 1.005-0.287)(1000) (499.97-303)=82.11J

 

SInce work is added in the system for compression

U=Q+w

Q=U-W

 

q= 82.11 - p2v2-p1v1/n-1 = 89.11-109.40=-0.027KJ (ANS)

[studiot]

 

q= 82.11 - p2v2-p1v1/n-1 = 89.11-109.40=-0.027KJ (ANS)

 

 

Are you sure of your sign conventions and arithmetic?

 

Does the negative sign mean heat is also added or evolved?

 

 

q= 82.11 - p2v2-p1v1/n-1 = 89.11-109.40=-0.027KJ (ANS)

 

 

?

Edited November 18, 2013 by studiot

[BOOMOOMOO]

Im not exactly sure myself,

 

But if i was to follow the equation.

 

if work is done on system, q= u - w

 

u =82.11

 

w=109.40

 

do advise.

 

thanks!

its a typo. its 82.11, not 89.11. sorry

[studiot]

 

T2= T1(p2/p1)^0.3/1.3 = 499.7k

 

Agreed

 

useful polytropic formulae are

[math]\frac{{{V_2}}}{{{V_1}}} = {\left( {\frac{{{P_1}}}{{{P_2}}}} \right)^{\frac{1}{n}}} = {\left( {\frac{{{T_1}}}{{{T_2}}}} \right)^{\frac{1}{{n - 1}}}}[/math]

 

 

Note that the pressure and temp ratios are the other way up to the volume ratio.

 

There are two versions of the First Law

 

It looks as if you are using the physicists and engineers version

 

[math]\Delta U = Q - W[/math]

 

That is delta U = Heat added to the system minus work done by the system.

 

Note that the first term on the right represents energy in and the second represents energy out.

 

Chemists use the alternative

 

[math]\Delta U = Q + W[/math]

 

In which we consider all forms of energy in as positive in out as negative.

 

For chemists delta U equals heat input to the sytem plus work done on the system.

Edited November 18, 2013 by studiot

[gabrelov]

That would be correct

Edited November 19, 2013 by gabrelov