Chemistry Work, temperature and energy in a constant pressure process

[Krokodrile]

Homework Statement

8 pounds of a gas with cp = 0.25 btu/lbm(ºF) and cv = 0.18 btu/lbm(ºF) that is at 100 ºF, is placed in a piston cylinder and 500 btu is delivered as heat in a no-flow process at constant pressure. Determine the final temperature, the internal energy change, and the work done

Relevant Equations

Q=m*C*△T
W = △U + Q

So, the Cp and Cv its very confusing for me. But, i understand what's its happening in this process, so, i use the logic and first i obtain a ecuation for obtain the final temperature ecuaticon:

Q=m*C*△T
Q=m*C*(T2-T1)
T2=(Q+T1)/(m*C)

If the process its in constant pressure, i use the Cp valor:

T2=(500 +100ºF)/(8 pounds * 0.25btu/lbm(ºF)
T2 = 300 ºF

So, i determinate the internal energy change using this ecuacion:

△U= 36 Cv*△T
△U = 0.18 btu/lbm(ºF) * (300 ºF - 100 ºF)
△U = 36 btu/lbm(ºF)

For the work:

W = △U + Q
W = 36 btu/lbm(ºF) `+ 500 btu
W = 536

I know that my results its incorrect, but these "Cp and Cv" confusing to me.

I need some help please :).

 

[Chestermiller]

You had the right idea, but in addition to understanding the physical principles, you need to do the math correctly. $$T_2=T_1+\frac{Q}{MC_p}=100+\frac{500}{(8)(0.25)}=350\ F$$
$$\Delta U=MC_v(350-100)=(8)(0.21)(250)=420\ BTU$$Your equation for the first law is incorrect. It should read $$\Delta U=Q-W$$where W is the work done by the system on its surroundings. $$W=Q-\Delta U=500-420=80\ BTU$$

 

[Krokodrile]

You had the right idea, but in addition to understanding the physical principles, you need to do the math correctly. $$T_2=T_1+\frac{Q}{MC_p}=100+\frac{500}{(8)(0.25)}=350\ F$$
$$\Delta U=MC_v(350-100)=(8)(0.21)(250)=420\ BTU$$Your equation for the first law is incorrect. It should read $$\Delta U=Q-W$$where W is the work done by the system on its surroundings. $$W=Q-\Delta U=500-420=80\ BTU$$

Ohhh, thank you so much. I see now where its my mistake, well ;) I am happy to understand by myself this problem.

 

[Krokodrile]

You had the right idea, but in addition to understanding the physical principles, you need to do the math correctly. $$T_2=T_1+\frac{Q}{MC_p}=100+\frac{500}{(8)(0.25)}=350\ F$$
$$\Delta U=MC_v(350-100)=(8)(0.21)(250)=420\ BTU$$Your equation for the first law is incorrect. It should read $$\Delta U=Q-W$$where W is the work done by the system on its surroundings. $$W=Q-\Delta U=500-420=80\ BTU$$

Excuse me Sir. In the class, when the teacher give us the answer of the problems, she use the △T for determinate the △U and W, so, the results was different:

△T = △H / m*Cp = 500 BTU/ (8lb) (0.25 BTU/lbmF) = 250 F

△U = m*Cv*△T = (8 lb) (0.18 BTU/lbmF) (250 F) = 360 BTU

W = Q - △U = 500 BTU - 360 BTU = 140 BTU

Its this correct? why she use △T and not T2?

The final temperatura its correct in both cases.

Thank you for read.

 

[Chestermiller]

Excuse me Sir. In the class, when the teacher give us the answer of the problems, she use the △T for determinate the △U and W, so, the results was different:

△T = △H / m*Cp = 500 BTU/ (8lb) (0.25 BTU/lbmF) = 250 F

△U = m*Cv*△T = (8 lb) (0.18 BTU/lbmF) (250 F) = 360 BTU

The version of Cv that I remember seeing was Cv=0.21. With 0.18, this is correct.

W = Q - △U = 500 BTU - 360 BTU = 140 BTU

Its this correct? why she use △T and not T2?

Why do you think she has not gotten W calculated correctly. It is correct.

The final temperatura its correct in both cases.

Of course the final temperature is correct in both cases. There is only one final temperature.

Thank you for read.