Pressure/temperature energy question

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Homework Statement



A cylinder contains 260 L of hydrogen gas (H2) at 0.0° C and a pressure of 10.0 atm. How much energy is required to raise the temperature of this gas to 28.0° C? give answer in kJ.

Homework Equations



C=Q/(delta)T and/or PV=nRT

The Attempt at a Solution



using the first equation I got an answer of 63.5 which is incorrect. I am not sure how to place the given values into the second equation because I don't know what R stands for.

the attempt with the second equation: (10.0 x 101.3)(260)=nR(28)
1013*260=nR28
263380=nR28
9406.428571 =nR
????????????

Thanks in advance.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
kuruman
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The second equation (ideal gas law) says nothing about energy, so it is inappropriate for this question. You must use the first equation, but I am not sure you have written correctly. Look up in you textbook what equation you must use when you add heat to a gas at constant volume.
 
  • #3
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The second equation (ideal gas law) says nothing about energy, so it is inappropriate for this question. You must use the first equation, but I am not sure you have written correctly. Look up in you textbook what equation you must use when you add heat to a gas at constant volume.
That explains why it made no sense.

I have another equation is it relevant?

U=Q + W
 
  • #4
kuruman
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More correctly ΔU = Q + W. That could be it because you are looking for Q. Note that the volume of the gas does not change, so W = 0 which means that ΔU = Q. Can you find an expression for ΔU?
 
  • #5
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I found the equation

(Delta)U = nC(sub V)*(Delta)T

I have changed the T into Kelvin and plugged in the following numbers:

(D)U = n(28.82)(301) = 8674.82

I have to give the answer in kJ. If the answer I got correct how do I put it into the right form? What is the conversion?
 
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  • #6
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I subsequently found another version of the equation as written above but with C(sub p). Then I found that:

Delta U= nC(sub p)*Delta T and worked out the problem as follows:
U = 12.5 (28)=350

Which one is the right one? Is there a right one?
 
  • #7
kuruman
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You need to know what these equations mean and what the symbols stand for. You use CV when you add heat heat to the gas at constant volume and Cp when you add heat at constant pressure. Here you are adding heat at constant volume.

In the expression Q = n CV ΔT

n = number of moles - you need to find that.
CV = specific heat at constant volume
ΔT = change in temperature (it is not 301)

When you put in all the correct numbers, the answer should come out in Joules. You can easily convert that to kJ at that point.
 
  • #8
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Q= n Cv delta T

n=3.348x10^-24 Cv= 10.144 T=48

Q= (3.348x10^-24)(10.144)(48)

Q= 1.630e^-21


this seems rather small. I think I'm not calculating the temperature correctly.
 
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  • #9
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Do I have to use the 10.0 atm someplace?
 
  • #10
kuruman
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Yes, this is extremely small. Please show exactly how you calculated n and the temperature change.
 
  • #11
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calculating n
molecular weight of hydrogen 2.016/6.022x10^23=3.34772501e^-24 I rounded this to 3.348

Cv=10.144 (this is from a chart but I think it is for 25 degree C) so there is probably an error here but I don't know how to adjust it)

Delta T=301-253=48
 
  • #12
kuruman
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calculating n
molecular weight of hydrogen 2.016/6.022x10^23=3.34772501e^-24 I rounded this to 3.348
The number of moles is proportional to the total mass of the gas. Your calculation shows that any mass of gas will have the same number of moles. This is not correct. Use the ideal gas law to find the amount of gas (number of moles) that you have

Cv=10.144 (this is from a chart but I think it is for 25 degree C) so there is probably an error here but I don't know how to adjust it)
The temperature correction is not to worry about. The units of this are kJ/kg K. This means that if you use this number, you need to find the mass of the gas from the number of moles and express it in kilograms. Remember that 6.022x1023 molecules are one mole (n = 1).

Delta T=301-253=48
0 oC = 273 K, not 253 K.
 
  • #13
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new calculations:

Q= n Cv T

n=(524.16) Cv=(10.144) Delta T=(28)

Q=148878.2131

To turn into kJoules = answer x 1000=148.878

this seem more plausible to me. What do you think?
 
  • #14
kuruman
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More plausible, yes, but you didn't show how you got the mass of the gas. CV=10.144 kJ/kg K. This means that if use this number, you need to convert the number of moles into kilograms. What is the meaning of n=524.16 and how did you get it?
 
  • #15
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More plausible, yes, but you didn't show how you got the mass of the gas. CV=10.144 kJ/kg K. This means that if use this number, you need to convert the number of moles into kilograms. What is the meaning of n=524.16 and how did you get it?
n=524.16 was found by multiplying the number of moles(2.016) by the 260L of hydrogen gas in the container from the original problem. I thought that might be how to find the value of n.

moles were found by multiplying the atomic mass of Hydrogen (1.0079) by the 2 atoms (Hydrogen gas is H sub2 right?) that make up H gas and rounding to 3 decimal places.

Thank you.
 
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  • #16
kuruman
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OK. You start with the ideal gas law pV = n R T.
The number of moles is represented by the symbol n, so
n = pV/RT

In the above expression p is in Pascal, V is in m3 and T is in Kelvin. You need to do some conversions here before you plug in.

Once you have a number for n, you know that one mole of hydrogen gas is 2 grams. Multiplying n by 2 gives you the mass of the gas in grams. You need to convert that to kilograms. Then you plug in the formula for Q.
 
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  • #17
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so then it should look like this:

n=pV/RT

n= (1.01325x10^6)(260)/R(28)


with p in Pascals; 260 being the liters of gas (or should I use the 10.144?); 28 is the change in temperature and R? is R the constant volume?

thanks. let me redo this. I somehow missed part of you message, sorry
 
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  • #18
kuruman
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with p in Pascals; 260 being the liters of gas (or should I use the 10.144?); 28 is the change in temperature and R? is R the constant volume?
The constant volume is represented by the symbol V (for volume) and in this case is 260 liters. It needs to be converted to cubic meters. It is not to be confused with the specific heat at constant volume CV.

R is the universal gas constant. It is equal to 8.31 J/K.mole.
 
  • #19
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OK. You start with the ideal gas law pV = n R T.
The number of moles is represented by the symbol n, so
n = pV/RT

In the above expression p is in Pascal, V is in m3 and T is in Kelvin. You need to do some conversions here before you plug in.

Once you have a number for n, you know that one mole of hydrogen gas is 2 grams. Multiplying n by 2 gives you the mass of the gas in grams. You need to convert that to kilograms. Then you plug in the formula for Q.
after converting I have

n=pV/RT

(1.01325x10^6)(260)/(10.144)(3.0115x10^2)=86237.83645=n

have I found n?
 
  • #20
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Ok so I now have ALL the conversions and a new answer of n=9547107.31
 
  • #21
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after dividing by 2 and converting to kg my answer is:

1.909421 x 10^4 kg

plugging in for Q then this is my final answer?
 
  • #22
kuruman
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after converting I have

n=pV/RT

(1.01325x10^6)(260)/(10.144)(3.0115x10^2)=86237.83645=n

have I found n?
Not correctly. I can see that p = 1.01325x10^6, that's correct.

260 is not V in cubic meters.

10.144 does not belong in this equation.

R does not show in the denominator.

I don't know what 3.0115x10^2 represents. If it is meant to be the temperature, it is not the temperature at which the pressure is 10 atm, i.e. it is not the initial temperature. The initial temperature is 273 K.
 
  • #23
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Not correctly. I can see that p = 1.01325x10^6, that's correct.

260 is not V in cubic meters.

10.144 does not belong in this equation.

R does not show in the denominator.

I don't know what 3.0115x10^2 represents. If it is meant to be the temperature, it is not the temperature at which the pressure is 10 atm, i.e. it is not the initial temperature. The initial temperature is 273 K.
oh ok then the new equation should look like this:

n=(1.01325x10^6)(2.6x10^-1) / (8.31)(273) =8654691.336

multiply by 2 = 17309382.67 grams

converted to kg= 1.730938267 x 10^7

and this number plug into Q=nCvT

right?
 
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  • #24
OmCheeto
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oh ok then the new equation should look like this:

n=(1.01325x10^6)(2.6x10^-1) / (8.31)(273) =8654691.336
I'm pretty sure you were supposed to divide by 273 instead of multiplying.
 
  • #25
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I'm pretty sure you were supposed to divide by 273 instead of multiplying.
in the equation

n= pV/RT

isn't pV divided by the quantity R times T?
 
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