When 1 mol gas is heated at constant volume, the temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. The correct statement among the following is:

1.  q = w = 500 J, ∆U = 0
2.  q = ∆U = 500 J, w = 0
3.  q =0, w = 500 J, ∆U = 0
4.  ∆U = 0, q = w = – 500 J

Identify the correct statement regarding entropy:

Two litres of an ideal gas at a pressure of 10 atm expands isothermally at 25 °C into a vacuum until its total volume is 10 litres. The amount of heat absorbed during expansion is:

1. 80 J

2. -80 J

3. Zero

4. 50 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:

A swimmer coming out from a pool is covered with a film of water weighing about 18g. The internal energy of vaporization at 298K. is-

∆_vap H^⊖ for water at 298K= 44.01kJ mol^–1

1. 38.63 kJ
2. 43.82 J
3. 41.53 kJ
4. 40.33 J

The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, CO₂(g) and H₂O (l) 
are produced and 3267.0 kJ of heat is liberated.
The standard enthalpy of formation, ∆_fH^⊖ of benzene is:
(Standard enthalpies of formation of CO₂(g) and ${\mathrm{H}}_2\mathrm{O}<mspace\; width="1em"></mspace>\left(\mathrm{l}\right)$ are –393.5 kJ mol^–1 and – 285.83 kJ mol^–1 respectively.)

1. 54. 24 kJ mol^–1
2. 48. 51 kJ mol^–1
3. 66. 11 kJ mol^–1
4. 15. 21 kJ mol^–1

For the oxidation of iron,

4Fe(s) + 3O₂ (g) →  2Fe₂O₃ (s)
entropy change is -549.4 JK^-1 mol^-1 at 298 K.

(${∆}_r{H}^{\Theta }$ for this reaction is $-1648<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>{10}^3<mspace\; width="1em"></mspace>J<mspace\; width="1em"></mspace>mo{l}^{-1}$)

The reaction is:

1. Spontaneous reaction

2. Non-spontaneous reaction

3. Reaction is an equilibrium

4. Cannot predict

Given
\(\begin{aligned} &\mathrm{{C}_{{(graphite) }}+{O}_{2}({~g})} → \mathrm{{CO}_{2}({~g})} \\ &\mathrm{\Delta_{r} {H}^{\circ}=-393.5 {~kJ} {~mol}^{-1}} \\ &\mathrm{H_{2}(g) + \frac{1}{2} {O}_{2}({~g})} → \mathrm{{H}_{2} {O}({l})} \\ &\mathrm{\Delta_{r} {H}^{\circ}=-285.8 {~kJ} {~mol}^{-1}} \\ &\mathrm{{CO}_{2}({~g})+2 {H}_{2} {O}({l})} → \mathrm{{CH}_{4}({~g})+2 {O}_{2}({~g})} \\ &\mathrm{\Delta_{r} {H}^{\circ}=+890.3 {~kJ} {~mol}^{-1}} \end{aligned}\)

Based on the above thermochemical equations, the value of Δ_rH° at 298 K for the reaction
\(\mathrm{C_{(graphite)} + 2 H_{2} (g) → CH_{4} (g)}\) will  be :

1. –74.8 kJ mol^–1

2. –144.0 kJ mol^–1

3. +74.8 kJ mol^–1

4. +144.0 kJ mol^–1

If for a certain reaction ${∆}_{r}H$ is 30 kJ mol^–1 at 450 K, the value of ${∆}_{r}S$ (in JK^–1 mol^–1) for which the same reaction will be spontaneous at the same temperature is: