1
AIPMT 2003
MCQ (Single Correct Answer)
+4
-1
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
2
AIPMT 2003
MCQ (Single Correct Answer)
+4
-1
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
3
AIPMT 2003
MCQ (Single Correct Answer)
+4
-1
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
4
AIPMT 2003
MCQ (Single Correct Answer)
+4
-1
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