2 mole of an ideal gas at 27ºC temp. is expanded reversibly from 2 lit. to 20 lit. Find entropy change (R = 2 cal/mol K): 

1. 92.1

2. 0

3. 4

4. 9.2

Heat of combustion ∆Hº for C(s), H₂(g) and CH₄(g) are – 94, – 68 and – 213 Kcal/mol. ∆Hº for C(s) + 2H₂(g) → CH₄ (g) is:

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

The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm^–3, respectively. If the standard free energy difference (∆Gº) is equal to 1895 J mol^–1, the pressure at which graphite will be transformed into diamond at 298 K is:

1. 11.08$\times {10}^8$ $\mathrm{Pa}$

2. $9.92\times {10}^7$ $\mathrm{Pa}$

3. $9.92\times {10}^6$ $\mathrm{Pa}$

4. 11.08$\times {10}^5$ $\mathrm{Pa}$

What is the entropy change (in JK^–1 mol^–1) when one mole of ice is converted into water at 0 ºC? (The enthalpy change for the conversion of ice to liquid water is 6.0 KJ mol^–1 at 0 ºC)

The formation of a solution from two components can be considered as:

The solution so formed will be ideal if:

1. ∆H_Soln = ∆H₁ + ∆H₂ + ∆H₃

2. ∆H_Soln = ∆H₁ + ∆H₂ – ∆H₃

3. ∆H_Soln = ∆H₁ – ∆H₂ – ∆H₃

4. ∆H_Soln = ∆H₃ – ∆H₁ – ∆H₂

The molar heat capacity of water at constant pressure, C, is 75 JK^–1 mol^–1. When 1.0 kJ of heat is supplied to 100 g of water which is free to expand, the increase in temperature of the water is:

$\frac{1}{2}{N_2}_{<mspace\; width="1em"></mspace>\left(\mathrm{g}\right)}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\frac{1}{2}{O}_{2<mspace\; width="1em"></mspace>\left(\mathrm{g}\right)}<mspace\; width="1em"></mspace>\to <mspace\; width="1em"></mspace>N{O}_{\left(\mathrm{g}\right)}<mspace\; width="1em"></mspace>;$
${∆}_{\mathrm{r}}H°<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>90<mspace\; width="1em"></mspace>kJ<mspace\; width="1em"></mspace>mo{l}^{-1}$
$N{O}_{\mathit{\left(}g\mathit{\right)}}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\frac{1}{2}{O_2}_{\mathit{\left(}g\mathit{\right)}}<mspace\; width="1em"></mspace>\to <mspace\; width="1em"></mspace>N{O_{\mathit{2}}}_{\mathit{\left(}g\mathit{\right)}}\hspace{0.17em};<mspace\; width="1em"></mspace>$
${∆}_{\mathrm{r}}\mathrm{H}°<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>-74<mspace\; width="1em"></mspace>\mathrm{kJ}<mspace\; width="1em"></mspace>{\mathrm{mol}}^{-1}$

The thermodynamic stability of NO_(g) based on the above data is:

1. Less than NO_2(g) 

2. More than NO_2(g)

3. Equal to NO_2(g)

4. Insufficient data

The entropy change in the surroundings when 1.00 mol of H₂O_(l) is formed under standard conditions is:

∆_fH^θ = –286 kJ mol^–1 

1. 952.5 J mol^-1

2. $979.7$ $\mathrm{J}$ ${\mathrm{mol}}^{-1}$

3. $949.7$ $\mathrm{J}$ ${\mathrm{mol}}^{-1}$

4. $959.7$ $\mathrm{J}$ ${\mathrm{mol}}^{-1}$

For the graph given below, it can be concluded that work done during the process shown will be-