Internal Energy of an Expanding Gas

AI Thread Summary
The discussion centers on calculating the work done, change in internal energy, and heat transfer during the expansion of a monatomic ideal gas. The work done by the gas during expansion is calculated using the formula w = PΔV, resulting in 3600 J. For the change in internal energy, the formula U = (3/2)nRT is suggested, emphasizing the need to find the difference between the final and initial internal energy states. The first law of thermodynamics is highlighted to clarify the relationship between internal energy, work, and heat. The participant expresses gratitude for the clarification on using the correct formulas.
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Homework Statement


2 moles of a monatomic ideal gas expands from state A to state B. PA = PB = 600 N/m2, VA = 3.0 m3 and VB = 9.0 m3.

i. Calculate the work done by the gas as it expands.
ii. Calculate the change in internal energy of the gas as it expands.
iii. Calculate the heat added to or removed from the gas during this expansion.

Homework Equations


ΔE = q-w
w = PΔV
PV = nRT
R = 8.314 J/mol*K

The Attempt at a Solution


i. w = PΔV
w = (600)(9-6) = 3600 J

ii. 4.186 J of work = cal of heat ?
3600/4.186 = 860 cals

I'm having troubles with ii and iii, I'm not sure if the converting of the work into calories is correct. I've looked at all the possible formulas and I'm not sure how to go about finding internal energy with only pressure and volume. Help, please!
 
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Have you come across the formula:
U = \frac{3}{2}nRT​
for an ideal gas? This should be found in any standard textbook.
So you just have to find U_{2} - U_{1} for the change in internal energy.

No, mechanical work is not equal to heat! They are separate concepts. You have to use the first law of thermodynamics
\Delta U = W + Q​
where \Delta U = change in internal energy, W = work done on the gas, Q = heat supplied to the system.
I see that you have already written a variant of the equation under "relevant equations". You just have to use that equation and substitute the relevant values.
 
Ahhh, I haven't seen that formula!
I've got it now, thank you.
 
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