2 Marks Questions

Question 12 Marks

Calculate the number of atoms in 39.4g gold. Molar mass of gold is $197g ~mole^{–1}$.

Answer

197gm gold has number of atoms = $6.023 \times 10^{23}$
1gm gold will have number of gold atoms $=\frac{6.023\times10^{23}}{197}$
39.4gm gold has number of Au atoms $=\frac{39.4\times6.023\times10^{23}}{197}$
= $1.2 \times 10^{23}$ atoms

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Question 22 Marks

The volume of a given mass of a gas at 27°C, 1 atm is 100cc. What will be its volume at 327°C?

Answer

By gas equation of ideal gas $P_1=1 \mathrm{~atm}, P_2=1 \mathrm{~atm}, V_1=100 \mathrm{cc}, V_2=? T_1=273+27=300 \mathrm{~K} \mathrm{~T}_2=327+273=600 \mathrm{~K}$
$\therefore\frac{\text{P}_1\text{V}_1}{\text{T}_1}=\frac{\text{P}_2\text{V}_2}{\text{T}_2}$
$\frac{1\times100}{300}=\frac{1\times\text{V}_2}{600}$
$\Rightarrow\text{V}_2=\frac{100\times600}{300}=200\text{cc}$
Units of $(P_1P_2)$ and $(V_1V_2)$ must be same separatery but unit of T must be in only on Kelvin scale.

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Question 32 Marks

Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters 1 A and 2 A. The gases may be considered under identical conditions of temperature, pressure and volume.

Answer

Mean free path$\lambda=\frac{1}{\sqrt{2}\pi\text{d}^2\text{n}}$
N = number of molecules per unit volume, d = diameter of moleculeas the condition for both gases are identiacal so n will be constant.
Or $\lambda\alpha\frac{1}{\text{d}^2}$
Or $\frac{\lambda_1}{\lambda_2}=\frac{\text{d}_2^2}{\text{d}_1^2}=\frac{(2)^2}{(1)^2}$
$\lambda_1:\lambda_2=4:1$

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