When 20 J of work done on a gas, 40 J of heat was released. If the initial internal energy of the gas was 70 J, what is the final internal energy?
Correct answer: A. 50J
- A. 50J
- B. 90J
- C. 60J
- D. 110J
Explanation
The problem is solved using the first law of thermodynamics, which states that the change in internal energy (∆U) is equal to the heat added to the system (Q) minus the work done by the system (W), expressed as ∆U = Q - W. In this scenario, 40 J of heat is released, making Q = -40 J, and 20 J of work is done on the gas, leading to W = -20 J. Hence, ∆U = -40 - (-20) = -20 J. The final internal energy, U2, is then calculated as U2 = ∆U + U1 = -20 + 70 = 50 J. This confirms that Option A is correct. Other options result from incorrect calculations or misunderstandings of the energy changes involved, leading to values that do not match the calculated final internal energy of 50 J.
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About First Law of Thermodynamics
The first law relates heat supplied, work done and the change in internal energy through energy conservation. Problems use sign conventions and apply the law to isothermal, adiabatic, isobaric and isochoric processes. Internal energy is a state function, while heat and work depend on the path followed.
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