1.50 moles of an ideal monatomic gas are initially at a temperature of 317 K. If the gas gains 2730 J of heat and performs 780 J of work, what is its final temperature?

Question

BlackTom AI · Accepted Answer

1. Relevant knowledge points and concepts
- First law of thermodynamics: ΔU = Q - W, where Q is heat added to the system and W is work done by the system (both positive here).
- For an ideal monatomic gas, internal energy U = (3/2) nRT, so ΔU = (3/2) nR ΔT.
- R (gas constant) = 8.314 J·mol⁻¹·K⁻¹.

2. Complete reasoning process step by step
Apply the first law to relate heat, work, and change in internal energy:
ΔU = Q - W.
For a monatomic ideal gas, set ΔU = (3/2) nR ΔT, so:
(3/2) nR ΔT = Q - W.
Solve for ΔT and then for final temperature Tf = Ti + ΔT.

3. Calculations with intermediate steps
Given: n = 1.50 mol, Ti = 317 K, Q = +2730 J, W = +780 J.
Compute Q - W:
Q - W = 2730 J - 780 J = 1950 J.
Compute nR:
nR = 1.50 mol × 8.314 J·mol⁻¹·K⁻¹ = 12.471 J·K⁻¹.
Use (3/2) nR ΔT = 1950 J → ΔT = (2/3) · (1950 J) / (nR).
Calculate:
1950 / 12.471 ≈ 156.429 K,
ΔT = (2/3) × 156.429 ≈ 104.286 K.
Final temperature:
Tf = Ti + ΔT = 317 K + 104.286 K ≈ 421.286 K.
Rounded to three significant figures: 421 K.

4. Why the correct answer is correct
Using the correct energy balance and the correct molar heat capacity for a monatomic ideal gas (Cv = 3/2 R) yields a temperature increase of about 104.3 K, producing a final temperature near 421 K. Sign conventions (Q positive in, W positive out) were applied correctly in ΔU = Q - W.

5. Why other options are incorrect
- 317 K: implies no net change in internal energy (would require Q = W), which is false here (2730 ≠ 780).
- 687 K: would correspond to a much larger ΔT; likely from misapplying formula or using an incorrect Cv.
- 359 K: too small an increase — could result from using Cv for a diatomic gas (5/2 R) or arithmetic error.
- none of these: not true because a correct calculation yields one of the listed choices.

6. Conclusion
The correct final temperature is 421 K, which matches the provided answer.

1.50 moles of an ideal monatomic gas are initially at a temperature of 317 K. If the gas gains 2730 J of heat and performs 780 J of work, what is its final temperature?单项选择题

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