Thermodynamic Query: Calculating Heat Transfer for Monoatomic Gas at S.T.P.

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The discussion revolves around calculating the net heat transfer for a monoatomic gas under specific conditions at standard temperature and pressure (S.T.P.). Initially, the temperature of one gram molecule of gas is raised by 1K at constant volume, followed by the addition of another gram molecule isothermally. Participants express confusion over the heat transfer calculations, particularly regarding work done during the isothermal process and the lack of clarity on how the gases are combined. The final consensus suggests that the net heat transfer for the entire process is 3 times the specific heat at constant volume. The problem is noted to be poorly defined, lacking details on the combination of gases.
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one gram molecule of monoatomic gas is taken at s.t.p. keeping its volume constant,its temperature is raised by 1k. one gram molecule of the gas is added isothermally to this.the mixture is now cooled by 1k at constant volume to return to s.t.p. if [c][/v] be gram molecular specific heat of gas constant volume, net heat transfer in the whole process is??

my solution: for 1st step- w=o,q=u=n[c][/v]dt=[c][/v]
2nd step-i don't know what will be q=?..as u=0, q=w, but how to find w?
3rd step-q=2[c][/v]
 
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What is the volume occupied by one gram molecule of a gas at s.t.p?
When you add some more gas, what happens to P, V, T? (T is constant since process is isothermal).
Once you have have found final volume, then use ##W=\int PdV##
 
This is a very poorly defined problem. You are not told how the two gases are combined. It could be at constant pressure, for all we know. That is the simplest case to consider.

Chet
 
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AdityaDev said:
What is the volume occupied by one gram molecule of a gas at s.t.p?
When you add some more gas, what happens to P, V, T? (T is constant since process is isothermal).
Once you have have found final volume, then use ##W=\int PdV##

volume at stp. is 22.4l. since we add 1mole gas it becomes 44.8l then should i use w=nrtlogv2/v1 (work done is isothermal process).
but if i use this answer is not coming.
i also thought that q=msdt and since temperature in second step is constant so q=0. is this correct??

the answer given is 3[c][/v].
 
Chestermiller said:
This is a very poorly defined problem. You are not told how the two gases are combined. It could be at constant pressure, for all we know. That is the simplest case to consider.

Chet

i am sorry but this is how the question was given in my book
 
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