Q. No.Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas? (Assume non-expansion work is zero)
1. Adiabatic process :
2. Cyclic process:
3. Isothermal process:
4. Isochoric process:
1. Adiabatic process :
2. Cyclic process:
3. Isothermal process:
4. Isochoric process:
What is the change in internal energy for 5 moles of an ideal gas when it undergoes reversible compression from 100 K to 200 K?
[Given CV = 28 J K–1 mol–1]
| 1. | ΔU = 8kJ | 2. | ΔU = 14kJ |
| 3. | ΔU = 10kJ | 4. | ΔU = 2.8 kJ |
What is the difference between \(\Delta H\) and \(\Delta U\) \((\Delta H - \Delta U)\) when the combustion of one mole of heptane(l) is carried out at a temperature T?
| 1. | 3RT | 2. | 4RT |
| 3. | – 3RT | 4. | – 4RT |
The change in internal energy, ∆U, in kJ, when a spring undergoes compression, with 10 kJ of work performed and 2 kJ escaping as heat to the surroundings, is:
| 1. | 12 | 2. | –8 |
| 3. | 8 | 4. | –12 |
Enthalpy of sublimation of iodine is 24 cal g–1 at 200 °C. If specific heat of I2(s) and l2 (vap) are 0.055 and 0.031 cal g–1K–1 respectively, then enthalpy of sublimation of iodine at 250°C in cal g–1 is :
1. 2.85
2. 22.8
3. 11.4
4. 5.7
An ideal gas is allowed to expand form 1 L to 10 L against a constant external pressure of 1 bar. The work done in kJ is:
1. +10.0
2. – 9.0
3. – 2.0
4. –0.9
The internal energy change (in kJ) when 90g of water undergoes complete evaporation at 100ºC is:
(Given : Hvap for water at 373 K = 41 kJ/mol, R = 8.314 JK–1mol–1)
1. 154.5
2. 168.5
3. 176.5
4. 189.5
(R = 2 cal mol–1 K–1)
1. 316 kcal
2. 326 kcal
3. 365 kcal
4. 342 kcal
For one mole of an ideal gas, which of these statements must be true?
(I) U and H each depend only on temperature.
(II) Compressibility factor z is not equal to 1.
(III) CP, m – CV, m = R
(IV) dU = CVdT for any process.
1. (I), (III) and (IV)
2. (II), (III) and (IV)
3. (III) and (IV)
4. (I) and (III)
Five moles of an ideal gas at 1 bar and 298 K are expanded into a vacuum till the volume doubles. The work done is:
1. –RT ln V2/V1
2. CV(T2 – T1)
3. zero
4. – RT(V2 – V1)
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