2 Marks Questions

Question 512 Marks

Explain the phenomenon of evaporation on the basis of kinetic theory.

Answer

According to kinetic theory, molecules of a liquid are in a state of continuous random motion. The molecules near the surface of liquid may have enough K.E. so as to overcome the intermolecular attraction of other molecules on the surface and hence manage to escape. Such molecules would move around freely in the space above the liquid. This is the phenomenon of evaporation which may occur at all temperatures.

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Question 522 Marks

Find the ratio of the mean speed of hydrogen molecules to the mean speed of nitrogen molecules in a sample contammg a mixture of the two gases.

Answer

$\text{V}_\text{avg}=\sqrt{\frac{8\text{RT}}{\pi\text{M}}}$$\frac{\text{V}_\text{avg}\text{H}_2}{\text{V}_\text{avg}\text{N}_2}=\sqrt{\frac{8\text{RT}}{\pi\times2}}=\sqrt{\frac{\pi\times28}{8\text{RT}}}=\sqrt{\frac{28}{2}}=\sqrt{14}=3.74$

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Question 532 Marks

Shows plot of $\frac{P V}{T}$ versus $P$ for $1.00 \times 10^{-3} \mathrm{~kg}$ of oxygen gas at two different temperatures: Which is true, $T_1>T_2$ or $T_1<T_2$ ?

Answer

The dotted plot in the given graph represents an ideal gas. The curve of the gas at temperature $T_1$ is closer to the dotted plot than the curve of the gas at temperature $T_2$. A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore, $\mathrm{T}_1>\mathrm{T}_2$ is true for the given plot.

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Question 542 Marks

Figure. shows a cylindrical tube of cross-sectional area A fitted with two frictionless pistons. The pistons are connected to each other by a metallic wire. Initially, the temperature of the gas is $T_0$ and its pressure is P, which equals the atmospheric pressure. (a) What is the tension in the wire? (b) What will be the tension if the temperature is increased to $2T_0$?

Answer

$\frac{\text{P}_1\text{V}_1}{\text{T}_1}=\frac{\text{P}_2\text{V}_2}{\text{T}_2}$$\Rightarrow\frac{\text{P}_0\text{V}}{\text{T}_0}=\frac{\text{P}'\text{V}}{2\text{T}_0}\Rightarrow\text{P}'=2\text{P}_0$
Net pressure = $P_0$ outwards
$\therefore$ Tension in wire = $P_0A$
Where A is area of tube.

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Question 552 Marks

Calculate the molecular K.E. of 1 gram of Helium (Molecular weight 4) at $127^\circ C$. Given $R = 8.31J mol^{-1} K^{-1}$.

Answer

T = 127 + 273 = 400K Average K.E. per mole of Helium $=\frac{3}{2}\text{RT}$ Average K.E. of I gram of Helium$=\frac{3}{2}\frac{\text{RT}}{\text{M}}=\frac{3\times8.31\times400}{2\times4}$
$=1246.5\text{J}$

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Question 562 Marks

The ratio of specific heat capacity at constant pressure to the specific heat capacity at constant volume of a diatomic gas decreases with increase in temperature. Explain, why.

Answer

We know $\gamma=\frac{\text{C}_\text{p}}{\text{C}_{\text{v}}}=1+\frac{2}{\text{f}},$ where f is the degree of freedom of a diatomic gas. The degree of freedom of a diatomic gas increases with the increase in temperature, so γ decreases with increase in temperature.

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Question 572 Marks

Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Given the radius of oxygen molecule as $3\mathring{\text{A}}.$

Answer

Radius of oxygen $=3\mathring{\text{A}}=3\times10^{-10}\text{m}$ Volume of each oxygen molecule $=\frac{4}{3}\pi\text{r}^3$$=\frac{4}{3}\times3.14\times27\times10^{-30}$
Molar volume at STP $=2.24\times10^4\text{m}^3,$$\frac{\text{Molecular volume}}{\text{Volume of gas}}=\frac{6.023\times10^{23}\times\frac{4}{3}\pi\text{r}^3}{2.24\times10^4}$
$=\frac{680.8\times10^{-7}}{2.24\times10^4}$
$=304\times10^{-11}$

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Question 582 Marks

A cylinder of fixed capacity 44.8 litres contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by $15.0^{\circ} C ?\left(R=8.31 J mol ^{-1} K ^{-1}\right)$.

Answer

Using the gas law $P V=\mu R T$, you can easily show that $1 mol$ of any $($ ideal $)$ gas at standard temperature $(273 K )$ and pressure $\left(1 atm =1.01 \times 10^5 Pa \right)$ occupies a volume of $22.4$ litres. This universal volume is called molar volume. Thus the cylinder in this example contains $2 mol$ of helium. Further, since helium is monatomic, its predicted $($ and observed $)$ molar specific heat at constant volume, $C_V=(3 / 2) R$, and molar specific heat at constant pressure, $C_p=(3 / 2) R+R=(5 / 2) R$. Since the volume of the cylinder is fixed, the heat required is determined by $C_V$. Therefore,
Heat required $=$ no. of moles $\times$ molar specific heat rise in temperature
$=2 \times 1.5 R \times 15.0=45 R$
$=45 \times 8.31=374 J .$

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Physics 2 Marks Questions