NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

### 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
2

### AIPMT 2003

For the reaction,
C3H8(g) + 5O2(g) $$\to$$ 3CO2(g) + 4H2O(l)
at constant temperature, $$\Delta$$H $$-$$ $$\Delta$$E is
A
+ RT
B
$$-$$3RT
C
+ 3RT
D
$$-$$ RT

## Explanation

$$\Delta$$H = $$\Delta$$E + $$\Delta$$ng RT

$$\Delta$$ng = 3 – (1 + 5) = – 3

$$\Rightarrow$$ $$\Delta$$H = $$\Delta$$E + (– 3RT)

$$\Rightarrow$$ $$\Delta$$H – $$\Delta$$E = – 3RT
3

### AIPMT 2002

2 mole of ideal gas at 27oC temperature is expanded reversibly from 2 lit. to 20 lit. Find entropy change. (R = 2 cal/mol K)
A
92.1
B
0
C
4
D
9.2

## Explanation

For isothermal reversible expansion

w = q = nRT $$\times$$ 2.303 $$\log {{{v_2}} \over {{v_1}}}$$

= 2RT $$\times$$ 2.303 $$\log {{{20}} \over {{2}}}$$

= 2 × 2 × T × 2.303 × 1 = 9.2 T

Entropy change, $$\Delta$$S = $${q \over T}$$ = $${{9.2T} \over T}$$ = 9.2 cal/mol K
4

### AIPMT 2002

Heat of combustion $$\Delta$$Ho for C(s), H2(g) and CH4(g) are $$-$$ 94, $$-$$ 68 and $$-$$213 kcal/mol, then $$\Delta$$Ho for C(s) + 2H2(g) $$\to$$ CH4(g) is
A
$$-$$17 kcal
B
$$-$$111 kcal
C
$$-$$170 kcal
D
$$-$$85 kcal

## Explanation

C(s) + 2H2(g) $$\to$$ CH4(g), $$\Delta$$Ho = ?

C(s) + O2(g) → CO2(g), $$\Delta$$H = – 94 kcal/mol ....(1)

H2(g) + $${1 \over 2}$$O2(g) $$\to$$ H2O(g), $$\Delta$$H = – 68 kcal/mol ....(2)

CH4(g) + 2O2(g) $$\to$$ CO2(g) + H2O(l), $$\Delta$$H = – 213 kcal/mol ....(3)

Performing (1) + 2 × (2) – (3)

$$\Delta$$Ho = [– 94 – 2(68)] – (–213) kcal/mol

= [– 94 – 136] + 213 k cal/mol

= – 230 + 213 = – 17 kcal/mol

### Joint Entrance Examination

JEE Main JEE Advanced WB JEE

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

NEET

Class 12