1

### AIPMT 2003

The molar heat capacity of water at constant pressure, C, is 75 J K$-$1 mol$-$1. When 1.0 kJ of heat is supplied to 100 g of water which is free to expand, the increase in temperature of water is
A
1.2 K
B
2.4 K
C
4.8 K
D
6.6 K

## Explanation

As we know, q = nC$\Delta$T

q = 1.0 kJ = 1000 J

C = 75 JK–1 mol–1

m = 100 g ⇒ Number of moles = ${{100} \over {18}}$ g mol–1

1000 = ${{100} \over {18}}$ $\times$ 75 $\times$ $\Delta$T

$\Rightarrow$ $\Delta$T = ${{10 \times 18} \over {75}}$ K

$\Rightarrow$ $\Delta$T = 2.4 K
2

### AIPMT 2003

Formation of solution from two components can be considered as
(i) Pure solvent $\to$ separated solvent molecules, $\Delta$H1
(ii) Pure solute $\to$ separated solute molecules, $\Delta$H2
(iii) Separated solvent and solute molecules $\to$ solution, $\Delta$H3
Solution so formed will be ideal if
A
$\Delta$Hsoln = $\Delta$H1 + $\Delta$H2 + $\Delta$H3
B
$\Delta$Hsoln = $\Delta$H1 + $\Delta$H2 $-$ $\Delta$H3
C
$\Delta$Hsoln = $\Delta$H1 $-$ $\Delta$H2 $-$ $\Delta$H3
D
$\Delta$Hsoln = $\Delta$H3 $-$ $\Delta$H1 $-$ $\Delta$H2

## Explanation

For ideal solution,

$\Delta$Hsoln = $\Delta$H1 + $\Delta$H2 + $\Delta$H3
3

### AIPMT 2003

For which one of the following equations is $\Delta$Horeact equal to $\Delta$Hof for the product ?
A
Xe(g) + 2F2(g) $\to$ XeF4(g)
B
2CO(g) + O2(g) $\to$ 2CO2(g)
C
N2(g) + O3(g) $\to$ N2O3(g)
D
CH4(g) + 2Cl2(g) $\to$ CH2Cl2(l) + 2HCl(g)

## Explanation

Heat of formation, $\Delta$Hof of a substance is the amount of heat absorbed or released when one mole of this substance is formed directly from its constituent elements.

In option (a), one mole of XeF4 is formed from its constituent elements i.e., Xe and F2 thus, the equation has equal value of $\Delta$Hor and $\Delta$Hof .

In option (b), the constituent atoms should be carbon and oxygen only but the reactant used is CO thus,

$\Delta$Hor $\ne$ $\Delta$Hof

In option (c), the reactant used is O3 which is again not in its element form thus,

$\Delta$Hor $\ne$ $\Delta$Hof

In option (d), two products are formed thus

$\Delta$Hor $\ne$ $\Delta$Hof
4

### AIPMT 2003

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 $\left( {\Delta {G^o}} \right)$ is equal to 1895 J mol$-$1, the pressure at which graphite will be transformed into diamond at 298 K is
A
11.14 $\times$ 108 Pa
B
11.14 $\times$ 107 Pa
C
11.14 $\times$ 106 Pa
D
11.14 $\times$ 105 Pa

## Explanation

C(graphite) $\to$ C(diamond)

Volume of graphite = ${{12} \over {2.25}} = 5.33$ cm3 mol–1

Volume of diamond = ${{12} \over {3.31}} = 3.63$ cm3 mol–1

$\Delta$V = Vgraphite – Vdiamond

= 1.70 cm3 mol–1

= 1.70 × 10–3 L mol–1

So, $\Delta$Go = P$\Delta$V

$\Rightarrow$ 1895 J mol–1 = P(1.70 × 10–3 L mol–1)

$\Rightarrow$ 1114.70 × 103 J/L = P

$\Rightarrow$ ${{1114.7 \times {{10}^3}} \over {101.33}}$ = P

$\Rightarrow$ P = 11000.69 atm × 1.013 × 105 Pa

= 11143.69 × 105 Pa

= 11.14 × 108 Pa