CBSE Class 11 Chemistry Exemplar Chapter 5: States of Matter NCERT Solutions

The observation is due to the lower atmospheric pressure at high altitudes like Shimla. Atmospheric pressure decreases as altitude increases. Water boils when its vapor pressure equals the surrounding atmospheric pressure. At lower atmospheric pressure, water boils at a temperature below 100°C. Consequently, food cooked in water at this lower temperature takes longer to cook. In contrast, a pressure cooker increases the pressure inside, raising the boiling point of water and thus cooking food faster.

The property of water that explains the spherical shape of rain droplets is surface tension. Surface tension is the tendency of liquid surfaces to shrink into the minimum surface area possible. For a given volume, a sphere has the minimum surface area. Therefore, water droplets, under the influence of surface tension, adopt a spherical shape to minimize their surface energy.

This question relates to Charles's Law, which states that at constant pressure, the volume of a given mass of a gas is directly proportional to its absolute temperature ($V\; \backslash propto\; T$ or $V\; =\; kT$). This is represented by a straight line passing through the origin in a V vs T graph. Boyle's Law states that at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure ($V\; \backslash propto\; \backslash frac\{1\}\{p\}$ or $pV\; =\; constant$).

From the graph, for a given temperature (represented vertically), the volume decreases as the pressure increases. We can see that $V\_1\; >\; V\_2\; >\; V\_3\; >\; V\_4$. Since volume is inversely proportional to pressure at constant temperature, a larger volume corresponds to a lower pressure.

Therefore, the order of pressures is .

Answer: (c)

London dispersion forces arise from temporary fluctuations in electron distribution, creating instantaneous dipoles that induce dipoles in neighboring particles. The strength of these forces depends on how easily the electron cloud of a particle can be distorted, which is known as polarisability. Larger and more complex molecules with more electrons are generally more polarisable, leading to stronger London dispersion forces. The magnitude of these forces is not directly dependent on the charge or mass of the particles, nor on the presence of permanent dipoles (which are characteristic of dipole-dipole forces).

Answer: (c) Polarisability of interacting particles

Dipole-dipole forces occur between polar molecules that have permanent dipoles. A permanent dipole arises from an unequal sharing of electrons in a covalent bond, creating a separation of charge. The ends of the dipole carry partial charges, denoted by δ+ and δ-. These partial charges are significantly smaller than the charge of a full electron or proton (unit electronic charge).

Answer: (c) Less than unit electronic charge

According to Dalton's Law of Partial Pressures, the partial pressure of a gas in a mixture is equal to the mole fraction of that gas multiplied by the total pressure of the mixture.

Given:\nRatio of moles of dihydrogen ($H\_2$) to dioxygen ($O\_2$) = 1:4.\nTotal pressure of the mixture = 1 atm.

Let the moles of $H\_2$ be $n\_\{H\_2\}\; =\; 1x$ and moles of $O\_2$ be $n\_\{O\_2\}\; =\; 4x$.\nTotal moles = $n\_\{H\_2\}\; +\; n\_\{O\_2\}\; =\; 1x\; +\; 4x\; =\; 5x$.

Mole fraction of dioxygen ($\backslash chi\_\{O\_2\}$) = $\backslash frac\{n\_\{O\_2\}\}\{n\_\{H\_2\}\; +\; n\_\{O\_2\}\}\; =\; \backslash frac\{4x\}\{5x\}\; =\; \backslash frac\{4\}\{5\}$.

Partial pressure of dioxygen ($P\_\{O\_2\}$) = $\backslash chi\_\{O\_2\}\; \backslash times\; \backslash text\{Total\; Pressure\}$

$P\_\{O\_2\}\; =\; \backslash frac\{4\}\{5\}\; \backslash times\; 1\; \backslash text\{\; atm\}\; =\; 0.8\; \backslash text\{\; atm\}$.

To convert this pressure to $Nm^\{-2\}$ (Pascals), we use the conversion factor 1 atm ≈ $1.01325\; \backslash times\; 10^5\; Nm^\{-2\}$.

$P\_\{O\_2\}\; =\; 0.8\; \backslash text\{\; atm\}\; \backslash times\; 1.01325\; \backslash times\; 10^5\; \backslash frac\{Nm^\{-2\}\}\{atm\}\; \backslash approx\; 0.8106\; \backslash times\; 10^5\; Nm^\{-2\}\; \backslash approx\; 8.1\; \backslash times\; 10^4\; Nm^\{-2\}$.

The closest option is $8\; \backslash times\; 10^4\; Nm^\{-2\}$.

Answer: (c) $8\; \backslash times\; 10^4\; \backslash text\{\; Nm\}^\{-2\}$