Thermodynamics question; internal energy and heat?

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Homework Statement


A helium balloon with a volume of 1m^3 and temperature of 27 deg c is placed in liquid nitrogen in an open container and cools to a temperature of 77 K. Assume that helium behaves like an ideal gas at constant atmospheric pressure and find the new volume of the gas in the balloon, the change in internal energy of the gas, the work done during the process, and the heat removed from the gas.

i got the new volume=.257m^3
and i got the work with constant pressure=73400J
but i can't get the internal energy and the heat

Homework Equations


Q=mcT
U=Q+W

The Attempt at a Solution


I attempted to find the number of moles and then find the mass to plug it into Q=mct to find Q, but I am not getting the correct answer. Please help
 

Answers and Replies

  • #2
vela
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I got a slightly different answer for the work done on the gas. Also, what are you using for the specific heat capacity when calculating Q?
 
  • #3
name_ask17
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i know the answers should be -185506J for the gas and U=-111506J but i don't know how to get there. i feel like i am doing this problem completley wrong.
 
  • #4
Andrew Mason
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i know the answers should be -185506J for the gas and U=-111506J but i don't know how to get there. i feel like i am doing this problem completley wrong.
Are you using [itex]Q = nC_p\Delta T[/itex]? What would you use for Cp?

Once you have determined the change in volume and the heat flow (out of the gas) you can determine the work done on the gas and then just apply the first law to find the change in internal energy.

You can also calculate the change in internal energy directly. What is the relationship between change in internal energy and change in temperature?

AM
 
  • #5
name_ask17
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Are you using [itex]Q = nC_p\Delta T[/itex]? What would you use for Cp
Would i use the specific heat of helium?
 
  • #6
Andrew Mason
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Would i use the specific heat of helium?
Use the specific heat at constant pressure: Cp = R + R(no. of degrees of freedom)/2

Btw, the answers you have given do not appear to be correct. You can see this is the case simply from the relationship between Q, W and U. Changes in Q, U and W are all negative. Therefore, the magnitude of Q must equal the magnitude of the change in U + magnitude of the work done on or by the gas, so |Q| > |U| and |Q|>|W|.

AM
 
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name_ask17
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what is R in the equation?
 
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  • #8
name_ask17
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and also when i do that, i have to find the mass to use mcT, so i used the volume divided by 22.4, then multiplied it by the molar mass of helium (4) to get .17857------is that what I am supposed to use for the mass?

also, these were the answers given by the instructor so they should be correct; we have to prove why they are correct.
 
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Andrew Mason
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  • #10
name_ask17
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im still confused because it did nto give me the right answer.
 
  • #11
Andrew Mason
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and also when i do that, i have to find the mass to use mcT, so i used the volume divided by 22.4, then multiplied it by the molar mass of helium (4) to get .17857------is that what I am supposed to use for the mass?

It is easier to use moles. Mass is irrelevant and using it simply complicates the algebra. First calculate V/T which, from the ideal gas law, we know is equal to the number of moles x R/P:

[tex]nR/P = V/T[/tex]

Since there has been no change in P, or n and R is a constant, the ratio of V/T is constant and so is nR = PV/T

[tex]\Delta Q = nC_p\Delta T = nR\frac{7}{2}\Delta T[/tex]

and

[tex]\Delta U = nC_v\Delta T = nR\frac{5}{2}\Delta T[/tex]

also, these were the answers given by the instructor so they should be correct; we have to prove why they are correct.
And if they aren't correct you have to show why they are not correct.

AM
 
  • #12
name_ask17
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alrighty. thanks
 
  • #13
Andrew Mason
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alrighty. thanks
I have to apologize. I read helium and thought H2 for some reason. Of course, this is a monatomic gas with Cv = 3/2 and Cp =5/2, so the answers given by your teacher ARE correct if [itex]\Delta Q = -185,506 J[/itex] and [itex]\Delta U = -111,506 J[/itex]. Sorry about any confusion I may have caused.

AM
 
  • #14
name_ask17
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That's okay. Thank you for clearing it up.
 
  • #15
name_ask17
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wait. so how am i supposed to find the heat removed? can you show me a formula because i tried everything from Q=mcT and Q=nCT? please help
 
  • #16
Andrew Mason
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wait. so how am i supposed to find the heat removed? can you show me a formula because i tried everything from Q=mcT and Q=nCT? please help
Since the process occurs at constant pressure, the heat lost is just:

[tex]\Delta Q = nC_p\Delta T[/tex]

where [itex]C_p = \frac{5}{2}R[/itex]

AM
 

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