1

### AIPMT 2004

The work done during the expansion of a gas from a volume of 4 dm3 to 6 dm3 against a constant external pressure of 3 atm is (1 L atm = 101.32 J)
A
$-$ 6 J
B
$-$ 608 J
C
+ 304 J
D
$-$ 304 J

## Explanation

Work done during the expansion, W = – pdV

W = –3 atm (6 dm3 – 4 dm3)

= – 3 atm ( 2 dm3 ) (1 dm3 = 1 L)

= – 3 atm × 2 L

= – 6 L atm

As, 1 L atm = 101.32 J

$\therefore$ W = – 6 × 101.32 J = – 607.92 J ≈ – 608 J
2

### 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
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

What is the entropy change (in J K$-$1 mol$-$1) when one mole of ice is converted into water at 0oC? (The enthalpy change for the conversion of ice to liquid water is 6.0 kJ mol$-$1 at 0oC).
A
20.13
B
2.013
C
2.198
D
21.98

## Explanation

H2O(s) → H2O(l)

$\Delta$H = 6.0 kJ mol–1

T = 0 + 273 K = 273 K; $\Delta$S = ?

$\Delta$S = ${{\Delta H} \over T}$ = ${{6000} \over {273}}$ = 21.98 JK–1 mol–1