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Calculate the amount of heat added to the ideal monatomic gas

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Tab Caps
#1
May5-13, 05:36 PM
P: 1
1. The problem statement, all variables and given/known data

Three moles of an ideal monatomic gas expand at a constant pressure of 3.00 atm ; the volume of the gas changes from 3.4010^-2 m^3 to 4.5010^-2 m^3

Calculate the amount of heat added to the gas


2. Relevant equations

pV = nRT
dQ=nCvdT
Cv = (3/2)R for a monatomic ideal gas

3. The attempt at a solution

Using pV = nRT I found that the initial temperature Ti = 414 K and Tf = 548 K.

Then I want to use the equation dQ = n(Cv)dT to solve for the amount of heat added in the system. The correct answer is 8360 J, but when I do:

3*(3/2)*R*(548-414) = 5010 J

I think my problem is coming from not understanding the difference between dT and [itex]\Delta[/itex]T, or dQ and [itex]\Delta[/itex] Q. I actually know Δ is just the change (final-initial), and the derivative is the infinitesimal rate of change, so perhaps I'm just trying the wrong formula since I don't have dT? Or is there a way to figure out dT from what I'm given?
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DrClaude
#2
May6-13, 04:05 AM
Sci Advisor
PF Gold
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P: 1,341
Quote Quote by Tab Caps View Post
I think my problem is coming from not understanding the difference between dT and [itex]\Delta[/itex]T, or dQ and [itex]\Delta[/itex] Q. I actually know Δ is just the change (final-initial), and the derivative is the infinitesimal rate of change, so perhaps I'm just trying the wrong formula since I don't have dT? Or is there a way to figure out dT from what I'm given?
Since nothing in the equation depends on ##T## or ##Q##, you can write
$$
\begin{array}{rcl}
dQ & = & n C_V dT \\
\int_{Q_i}^{Q_f} dQ & = & n C_V \int_{T_i}^{T_f} dT \\
\left. Q \right|_{Q_i}^{Q_f} & = & n C_V \left[ T \right]_{T_i}^{T_f} \\
\Delta Q & = & n C_V \Delta T
\end{array}
$$
So that is not your problem.

Look carefully at the equation. What does ##V## in ##C_V## stand for?
Chestermiller
#3
May6-13, 10:08 AM
Sci Advisor
HW Helper
Thanks
PF Gold
Chestermiller's Avatar
P: 5,172
Hi Tab Caps. Welcome to Physics Forums.

You need to apply the first law. How much work was done on the surroundings?

You already calculated the change in internal energy of the gas. According to the first law, the total heat added is equal to the change in internal energy plus the work done on the surroundings.

Dr Claude alluded to a better (equivalent) way of getting the same result, by recalling that, in a constant pressure process on a closed system, the heat added is equal to the change in enthalpy. For an ideal gas, do you remember how the enthalpy change is related to the temperature change?

Chet


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