Q. No.| Assertion (A): | The change in internal energy, ΔE, for the vaporisation of 1 mol of water at 1 atm and 373 K is zero. |
| Reason (R): | For every isothermal process, ΔE = 0. |
| 1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are True, but (R) is not the correct explanation of (A). |
| 3. | (A) is True and (R) is False. |
| 4. | Both (A) and (R) are False. |
| 1. | In a reversible process, the system and surroundings are always in equilibrium with each other. |
| 2. | Work done in free expansion > 0. |
| 3. | For adiabatic change, \(\Delta q \) = 0 . |
| 4. | For a process carried at constant pressure, \(\Delta H = q_p \) |
Which of the following reactions has the least difference between the change in enthalpy (∆H) and the change in internal energy (∆E) at a given temperature?
1. \(2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{SO}_3(\mathrm{~g})\)
2. \(\mathrm{CaCO}_3(s) \rightarrow \mathrm{CaO}(s)+\mathrm{CO}_2(g)\)
3. \(\mathrm{NH}_4 \mathrm{SH} (s) ~~\rightarrow ~\mathrm{NH}_3(g)+\mathrm{H}_2 \mathrm{~S}(g)\)
4. \(2 \mathrm{NH}_3 (g) ~\rightarrow ~\mathrm{N}_2(g)+3 \mathrm{H}_2(g)\)
The standard enthalpy of vaporization for water at 100 oC is 40.66 kJ mol–1. The internal energy of vaporization of water at 100 oC (in kJ mol–1) is:
(Assume water vapour behaves like an ideal gas.)
| 1. | +37.56 | 2. | –43.76 |
| 3. | +43.76 | 4. | +40.66 |
1. \(\Delta U=q_p\)
2. \(\Delta U=q_v \)
3. W<0
4. W>0
1. \(\mathrm{\Delta {H}+\Delta {U}=\Delta {nR}} \)
2. \(\mathrm{\Delta {H}=\Delta {U -\Delta n_gRT}}\)
3. \(\mathrm{\Delta {H}=\Delta {U+\Delta n_gRT }}\)
4. \(\mathrm{\Delta {H} -\Delta {U=-\Delta n_gRT}}\)
\(\mathrm{A}_2(\mathrm{~g})+3 \mathrm{~B}_2(\mathrm{~g}) \rightarrow 2 \mathrm{AB}_3(\mathrm{~g})\)
If at 300K, the enthalpy change for the above reaction is +15 kJ, then the internal energy change is:
| 1. | 19.98 K J | 2. | 200 J |
| 3. | 1999 J | 4. | 1.9988 kJ |
Find the value of the standard enthalpy of formation (ΔfH°).
1. Zero
2. Less than ΔfU°
3. Greater than ΔfU°
4. Equal to ΔfU°
| 1. | 0.00459 J | 2. | 0.0747 J |
| 3. | 1340 J | 4. | 4690 J |
If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1 bar and 100°C is 41kJ mol–1. The internal energy change, when 1 mol of water is vapourised at 1 bar pressure and 100°C is:
| 1. | 35.5 kJ mol–1 | 2. | 37.9 kJ mol–1 |
| 3. | 41 kJ mol–1 | 4. | 44.2 kJ mol–1 |
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