Solving a Monatomic Gas Cooling Problem

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To solve the problem of a monatomic gas cooled by 50°C at constant volume with 830 J of energy removed, the First Law of Thermodynamics is essential. Since the volume is constant, the work done (dW) is zero, leading to the conclusion that the change in heat (dQ) equals the change in internal energy (dU). For monatomic gases, the molar specific heat at constant volume (Cv) is 3/2 times the gas constant (R). By substituting these values into the equation dU = nCv(dT), the number of moles (n) can be calculated. Understanding these principles is crucial for solving thermodynamic problems involving monatomic gases.
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I really have no Idea where to start... Please help

Homework Statement


A monatomic gas is cooled by 50 C at constant volume by removing 830 J of energy. How many moles of the gas is in the sample.


Homework Equations



?Q=mc (delta)T



The Attempt at a Solution




I am not sure where to start...a monatomic gas could be any of the noble gases, right? each with a different molecular weight and each would have a different specific heat at a constant volume(c). Is this even the right equation to start with. Thanks.
 
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The main principle in this question is the First law of Thermodynamics
By 1st Law, dQ=dW+dU (using d for delta,nevertheless)
dW=0 as volume is constant
So, dQ=dU

Now, for monoatomic gases, Molar specific heat at constant volume is given by 3/2.R,where R=gas const.
as dU=nCdT, the values can be substituted to get the only unknown n, which is the no.of moles.

(Monoatomic is explicitly mentioned so that the Cv of the gas can be found out based on equipartition of energy)
 
Thanks

Thank you... That was perfect. It is frustrating because the formulas you used were never taught to me. They are in the book but they are outside the material we were required to learn. But my teacher is like that... Any how thanks for your help.
 
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