Question: Now Tony settles upon a Carnot engine. Calculate the efficiency of the Carnot engine if the temperature it operates between are 350 K and 250 K.
Options:
| A. | 27 | |
| B. | 57 | |
| C. | 37 | |
| D. | 47 | |
| E. | along the line CD |
Answer: Option A
: A
The efficiency of a Carnot engine is given by the formula ηH.E=1−TLTH Substituting for known values we get ηH.E=1−250350=27
: A
The efficiency of a Carnot engine is given by the formula ηH.E=1−TLTH Substituting for known values we get ηH.E=1−250350=27
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Question 1. An ideal gas in a heat engine executes the cycle shown. Where is the temperature of the gas maximum?
- along the line BC
- at point B
- at point C
- at point D
- along the line CD
Answer: Option C
: C
Temperature is maximum at point C. for ideal gas, pV = NRT, so T is max, when pV is max, which corresponds to the point furthest from the origin.
: C
Temperature is maximum at point C. for ideal gas, pV = NRT, so T is max, when pV is max, which corresponds to the point furthest from the origin.
Question 3. The figure here shows five paths traversed by a gas on a P-V diagram. Rank the paths according to the change in internal energy of the gas, greatest first.
- 5 > 4 = 3 > 2 = 1
- 5 > 4 = 3 > 1 = 2
- 4 = 3 > 5 > 1 = 2
- 5 > 4 = 3 = 2 = 1
- along the line CD
Answer: Option D
: D
For any ideal gas, the internal energy can be related to the temperature as - Eint=nCvT ⇒ΔEint=nCvΔT (for a change of state). Thus the change in internal energy is solely dictated by the change in temperature.Since, all the processes 1, 2, 3, and 4 begin and end at the same temperatures, their ΔT′s(and,ΔEint=nCv(T2−T1)) are the same. Process 5, on the other hand goes through a larger ΔT(T3−T1), hence gains the highest internal energy.
: D
For any ideal gas, the internal energy can be related to the temperature as - Eint=nCvT ⇒ΔEint=nCvΔT (for a change of state). Thus the change in internal energy is solely dictated by the change in temperature.Since, all the processes 1, 2, 3, and 4 begin and end at the same temperatures, their ΔT′s(and,ΔEint=nCv(T2−T1)) are the same. Process 5, on the other hand goes through a larger ΔT(T3−T1), hence gains the highest internal energy.
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