How Can I Determine the Volume of a Diatomic Ideal Gas in a Rigid Container?

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The discussion focuses on calculating the volume of a diatomic ideal gas in a rigid container subjected to heating. The initial state is defined at 300K and 1.3x10^5 Pa, with a final state of 330K after adding 623.25J of heat. The pressure after heating is calculated to be 1.43 x 10^5 Pa, with no work done by the gas due to constant volume. The change in internal energy equals the heat added, resulting in 623.25J. The specific heat at constant volume (Cv) is assumed to be 5R/2, where R is the universal gas constant (8.31 J/(mol·K)).

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A diatomic ideal gas is in a rigid container at an initial sate of 300K and 1.3x10^5 Pa is heated to a final state of 330K by 623.25J of heat. calculate the pressure in the container after the heating, the work done by the gas, the change in internal energy and the volume of the container. if the same gas is now continued in another cylinder and is compressed with a different compression process from the same initial to final states as above, calculate the change of the internal energy. The universal gas constant is given as 8.31.

I 've managed to do the first three parts of the question but I'm curently stuck at finding the volume.

I've tried using pV=nRT but it doesn't work since i do not know the number of moles of gas present.

Aside from finding the volume, i don't really understand what does the last part of the question mean. Could someone explain it to me?
 
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For an ideal gas, internal energy is solely a function of temperature. It is computed by the following:

dU = Cv * dT where Cv is the constant volume specific heat and dT is temperature change.

If you don't know the quantity of gas, then the change in internal energy will be on a per unit mass. Note that above equation does not have mass term so it is per unit mass.

Do you know what polytropic processes are where P*V**n = constant, where n takes on different values based on whether process is isothermal, isometric, isentropic, or isobaric?
 


LawrenceC said:
For an ideal gas, internal energy is solely a function of temperature. It is computed by the following:

dU = Cv * dT where Cv is the constant volume specific heat and dT is temperature change.

If you don't know the quantity of gas, then the change in internal energy will be on a per unit mass. Note that above equation does not have mass term so it is per unit mass.

Do you know what polytropic processes are where P*V**n = constant, where n takes on different values based on whether process is isothermal, isometric, isentropic, or isobaric?

But specific heat isn't given in the question?

i do know what isothermal, isometric and isobaric means but i do not know about the P*V**n = constant equation.
 


Please show how you did the first three parts of the problem.
 


P1/T1=P2/T2
1.3 X 10^5 /300 = P2/330
P2=1.43 X 10^5 Pa

No work is done by gas since the volume of the container do not change.

Change in internal energy = heat, W=0
Hence, Change in internal energy = 623.25J
 


Latios1314 said:
P1/T1=P2/T2
1.3 X 10^5 /300 = P2/330
P2=1.43 X 10^5 Pa

No work is done by gas since the volume of the container do not change.

Change in internal energy = heat, W=0
Hence, Change in internal energy = 623.25J
Try to work out the number of moles from the heat capacity of this gas. The heat capacity is 623.25/30 = 20.775 Joules/K = 2.5R = 5R/2

The author of the problem should tell you what the Cv of this gas is since it is not possible to determine the Cv of a non-specfied diatomic ideal gas purely from theory. However, I think the author wants you to assume that the specific heat at constant volume is 5R/2. If so, how many moles are we dealing with?

[Note: The diatomic gas has three degrees of translational freedom. There are also possible degrees of rotational freedom (2) and vibrational freedom (2). Generally, the vibrational mode of most diatomic molecules is not active at temperatures of 300K so the vibrational modes do not contribute to the specific heat. However, at 300K, most gases will have 2 rotational degrees of freedom. So most diatomic gases in this temperature range will have a Cv = 5R/2 and a Cp = 7R/2].

AM
 

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