Work
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moles of a perfect gas undergoes a cyclic process (see figure) consisting of the following processes:
: Isothermal expansion at a temperature so that the volume is doubled and pressure changes.
: Isobaric compression at pressure to initial volume.
: Isochoric change leading to change of pressure from.
Total work done in the complete cycle is –
- 1672.5 J
- 1728 J
- -1728 J
- -1572.5 J
What is value of W4?
- −50 J
- 100 J
- 150 J
- 50 J
What do you mean by internal energy?
- 100 Nm
- 200 Nm
- 150 Nm
- 250 Nm
- 1600 J
- 100 J
- 400 J
- 600 J
- √3qElm
- √qElm
- √2qElm
- √5qElm
- 540 cal
- 40 cal
- Zero cal
- 500 cal
- Temperature
- Pressure
- Volume
- All of these
A typical dorm room or bedroom contains about 2500 moles of air. Find the change in the internal energy of this much air when it is cooled from 35.0∘C to 26.0∘C at a constant pressure of 1.00 atm. Treat the air as an ideal gas with γ = 1.400.
4.68 × 105 J
5.14 × 104 J
−4.68 × 105 J
−5.14 × 104 J
- 750 J
- 775 J
- 765 J
- 790 J
A bubble of 5.00 mol of helium is submerged at a certain depth in liquid water when the water (and thus the helium) undergoes a temperature increase ΔT of 20.0∘C; at constant pressure. As a result, the bubble expands. The helium is monoatomic and ideal. How much work W is done by the helium as it expands against the pressure of the surrounding water during the temperature increase? Use the first law of thermodynamics.
820 J
831 J
883 J
805 J
- Temperature
- Pressure
- Volume
- All of these
- final pressure of the gas is Pa+kxos+mgs
- work done by the gas 12kx2o+mgxo
- Internal energy of the gas decreases
- all the above
- 1 J
- -1 J
- 2 J
- – 2 J
- 1672.5 J
- 1728 J
- -1728 J
- -1572.5 J
- 100 Nm
- 200 Nm
- 150 Nm
- 250 Nm
- nRT loge(V2−nβV1−nβ)+αn2(V1−V2V1V2)
- nRT log10(V2−αβV1−αβ)+αn2(V1−V2V1V2)
- nRT loge(V2−nαV1−nα)+βn2(V1−V2V1V2)
- nRT loge(V1−nβV2−nβ)+αn2(V1V2V1−V2)
- 1672.5 J
- 1728 J
- −1728 J
- −1572.5 J
- 15 kJ
- 10 kJ
- 2.5 kJ
- 12.5 kJ
- 4 P0V0
- 6 P0V0
- 4, 5 P0V0
- 2 P0V0
- 0.02 J
- 0.05 J
- 0.08 J
- 0.1 J
- Work done is maximum along AB
- Work done is minimum along AB
- Work done along ACB = work done along ADB
- Work done along ADB is minimum
What would be the work done by the gas if the entire process is reversed (3 - 2 - 1)?
-30 J
30 J
-22.5 J
22.5 J
- final pressure of the gas is Pa+kxos+mgs
- work done by the gas 12kx2o+mgxo
- Internal energy of the gas decreases
- all the above
- 1672.5 J
- 1728 J
- −1728 J
- −1572.5 J
[ Consider series expansion of ln 2 = (1−12+13−14… )]
- Zero
- −2kq2d ln2
- kq2d ln2
- −kq2d ln2
- 750 J
- 775 J
- 765 J
- 790 J
- always same in all frames of reference.
- always different in different frames of reference.
- can be different in different frames of reference.
- none of these.
- 40 cal
- 30 cal
- 16.7 cal
- 10 cal