How much work is done on a gas?

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SUMMARY

The discussion centers on calculating the work done on a monatomic ideal gas when 4390 J of heat flows out of the system, maintaining constant temperature. The correct application of the first law of thermodynamics is crucial, where the equation \(\Delta E = Q + W\) clarifies that work done on the system is negative when heat is lost. The final answer is determined to be W = 4390 Joules, indicating that the work done on the gas is equal to the heat lost.

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  • Understanding of the first law of thermodynamics
  • Familiarity with monatomic ideal gas properties
  • Knowledge of heat transfer concepts
  • Ability to apply sign conventions in thermodynamic equations
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  • Study the first law of thermodynamics in detail
  • Learn about monatomic ideal gas behavior and properties
  • Explore heat transfer mechanisms in thermodynamic systems
  • Review sign conventions in thermodynamic equations and their implications
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This discussion is beneficial for students studying thermodynamics, physics enthusiasts, and anyone seeking to understand the principles of energy conservation in gas systems.

Chele
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First of all...thanks for any help. This is ultrabasic introduction to therm, so I know everyone says that no question is stupid, but I just feel like I'm really missing the boat here.

Homework Statement



The temperature of a monatomic ideal gas remains constant during a process in which 4390 J of heat flows out of the gas. How much work (including the proper + or - sign) is done on the gas?

Homework Equations



\DeltaE=nCv\DeltaT

and I think...maybe

\DeltaE=Q - W

The Attempt at a Solution



Since it says that the temperature remains constant, I was thinking that \DeltaE=0

Q is the heat supplied TO the system, so if 4390 J of heat came out, then it would be -Q.

So I figured that:
\DeltaE=Q - W
0=-4390 - W
W=-4390 Joules

That's not correct. They must be telling me that it's a monatomic ideal gas for some reason.

HELP!
 
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Sign convention

Chele said:
\DeltaE=Q - W
Realize that in this equation:
Q = the heat energy transferred into the the system
W = the work done by the system (on the environment)

So the work done on the system is just the negative of that.

Many modern books use this form of the first law:
\DeltaE=Q + W, in which W is the work done on the system. This form makes it clearer that this is just a statement of energy conservation.
 
Thank you...obviously I was looking at the meanings incorrectly.

The answer is W=4390 Joules.

The way you worded your answer actually helps me with yet another problem I was having! Have a great day!
 

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