NEET States of Matter Previous Year Questions with Solutions

Intermolecular forces are the forces of attraction and repulsion between interacting particles (atoms and molecules). The term covers van der Waals forces and hydrogen bonding. As explicitly defined in standard textbooks, this term does not include electrostatic ionic bonds or the chemical forces that hold atoms together within a molecule, namely covalent bonds:

Analyzing the physical nature of each given interaction:

Because covalent bonding (D) is an intramolecular interaction, any valid option must include A, B, C, and E while omitting D.

Boyle’s Law states that at constant temperature (T) and constant mass of gas (n), the pressure of a given mass of gas is inversely proportional to its volume:

Rearranging the equation of state with y = P and x = 1/V yields a linear function:

Because slope m = nRT, the slope is directly proportional to absolute temperature:

Option (1) plots P vs T (Gay-Lussac’s isochores). Option (2) plots P vs V hyperbolas but inverts the temperature ordering. Option (4) shows a hyperbola for P vs 1/V, violating linearity. Graph (3) is uniquely correct:

Boyle's law states that at constant temperature $T$, the pressure of a fixed mass of gas is inversely proportional to its volume: $$P = \frac{k_1}{V} \implies PV = k_1 = nRT$$ A plot of $P$ versus $V$ at constant temperature gives a rectangular hyperbola called an isotherm.

The constant $k_1$ is directly proportional to the absolute temperature $T$. A higher temperature corresponds to a larger value of $k_1 = nRT$. For any given volume $V$ on the horizontal axis:

- Graph (1): Inverted temperature labels ($200\text{ K}$ is shown at top and $600\text{ K}$ at bottom), which violates thermodynamics. - Graph (2): Shows pressure independent of volume (horizontal line), which is false. - Graph (3): Shows linear increase ($P \propto V$), which is isobaric/Charles law plot type, not $P$ vs $V$. - Graph (4): Shows three hyperbolic isotherms with $600\text{ K}$ on top, $400\text{ K}$ in the middle, and $200\text{ K}$ at the bottom. This is the exact NCERT textbook plot.

Using molar masses $M(O_2) = 32\text{ g mol}^{-1}$ and $M(H_2) = 2\text{ g mol}^{-1}$: - Moles of oxygen: $$n_{\text{O}_2} = \frac{m_{\text{O}_2}}{M_{\text{O}_2}} = \frac{4\text{ g}}{32\text{ g mol}^{-1}} = \frac{1}{8} = 0.125\text{ mol}$$ - Moles of hydrogen: $$n_{\text{H}_2} = \frac{m_{\text{H}_2}}{M_{\text{H}_2}} = \frac{2\text{ g}}{2\text{ g mol}^{-1}} = 1.0\text{ mol}$$

Assuming no chemical reaction occurs at $0^\circ\text{C}$ in the absence of a spark or catalyst, Dalton's law states that each gas behaves independently: $$n_{\text{total}} = n_{\text{O}_2} + n_{\text{H}_2} = 0.125 + 1.0 = 1.125\text{ mol}$$

Given $V = 1.0\text{ L}$, $T = 273\text{ K}$, and $R = 0.082\text{ L atm mol}^{-1}\text{K}^{-1}$: $$P_{\text{total}} = \frac{n_{\text{total}} R T}{V} = \frac{1.125\text{ mol} \times 0.082\text{ L atm mol}^{-1}\text{K}^{-1} \times 273\text{ K}}{1.0\text{ L}}$$

Molar mass of diatomic nitrogen ($$\text{N}_2$$) = $$2 \times 14 = 28\text{ g mol}^{-1}$$. Molar mass of noble gas argon ($$\text{Ar}$$) = $$40\text{ g mol}^{-1}$$. $$n_{\text{N}_2} = \frac{7\text{ g}}{28\text{ g mol}^{-1}} = 0.25\text{ mol}$$ $$n_{\text{Ar}} = \frac{8\text{ g}}{40\text{ g mol}^{-1}} = 0.20\text{ mol}$$

$$n_{\text{total}} = n_{\text{N}_2} + n_{\text{Ar}} = 0.25 + 0.20 = 0.45\text{ mol}$$

$$x_{\text{N}_2} = \frac{n_{\text{N}_2}}{n_{\text{total}}} = \frac{0.25}{0.45} = \frac{25}{45} = \frac{5}{9}$$

According to Dalton's law, the partial pressure of a gas in a mixture equals its mole fraction multiplied by the total pressure: $$p_{\text{N}_2} = x_{\text{N}_2} \cdot P_{\text{total}}$$ $$p_{\text{N}_2} = \left(\frac{5}{9}\right) \times 27\text{ bar} = 5 \times 3 = 15\text{ bar}$$ (Notice also that for argon, $$p_{\text{Ar}} = \frac{4}{9} \times 27 = 12\text{ bar}$$, and $$15 + 12 = 27\text{ bar}$$).