Gaseous State — NEET UG practice

1. Start from the ideal gas equation, since it relates pressure, volume, and temperature to the amount of gas:

$$PV=nRT$$

2. Since the number of moles $n=\frac{w}{M}$ (mass divided by molar mass), substitute this in:

$$PV=\frac{w}{M}RT$$

3. Rearranging to get an expression for density $\rho=\frac{w}{V}$:

$$\rho=\frac{PM}{RT}$$

4. This tells us density is directly proportional to pressure and inversely proportional to temperature, since $M$ and $R$ are constants for a given gas.

So, to get the highest density, we need the option with the highest pressure and the lowest temperature.

5. Compare all four options:

• STP: $T=273\text{ K}$, $P=1\text{ atm}$
• 0°C, 2 atm: $T=273\text{ K}$, $P=2\text{ atm}$
• 0°C, 4 atm: $T=273\text{ K}$, $P=4\text{ atm}$
• 273°C, 4 atm: $T=546\text{ K}$, $P=4\text{ atm}$

6. Among these, 0°C, 4 atm has the highest pressure at the lowest possible temperature (273 K), since the last option has the same pressure but a much higher temperature (546 K), which lowers its density.

Hence, the answer is C.

1. Recall Boyle's law: for a fixed amount of gas at constant temperature, pressure is inversely proportional to volume:

$$P\propto\frac{1}{V}$$

2. Using the ideal gas equation $PV=nRT$, we can write pressure as a function of $\frac{1}{V}$:

$$P=(nRT)\left(\frac{1}{V}\right)$$

3. This is the equation of a straight line through the origin when $P$ is plotted against $\frac{1}{V}$, where the slope equals $nRT$.

4. Since temperature $T$ appears in the slope, a higher temperature gives a steeper slope (a line rising more sharply).

Therefore, if a graph shows multiple straight lines for different temperatures $T_1$, $T_2$, $T_3$, the line with the greatest slope corresponds to the highest temperature:

$$T_3>T_2>T_1$$

5. Option A correctly shows straight lines through the origin for $P$ vs $\frac{1}{V}$ with increasing slopes corresponding to increasing temperatures, matching Boyle's law.

Hence, the answer is A.

1. Recall that an ideal gas is assumed to have point-sized molecules with no attraction between them, but a real gas does not behave this way.

Because real molecules occupy some space and attract each other, the ideal gas equation $PV=RT$ needs two corrections.

2. Since real molecules have finite size, the volume actually available for movement is less than the measured volume $V$.

So we subtract a correction term $b$ (called the excluded volume) from $V$:

$$V_{available}=(V-b)$$

3. Since real molecules attract each other, the pressure they exert on the walls is slightly less than the ideal pressure.

Therefore a correction term $\frac{a}{V^2}$ is added to the observed pressure $p$ to make up for this loss:

$$p_{ideal}=\left(p+\frac{a}{V^2}\right)$$

4. Combining both corrections for 1 mole of gas gives the van der Waals equation:

$$\left(p+\frac{a}{V^2}\right)(V-b)=RT$$

5. Checking the options, only option A matches this exact form (with $n=1$, so no $n$ appears).

Option B wrongly places $V^2$ in the numerator instead of the denominator; option C has the wrong exponent on $a$ and mismatched volume term; option D uses $n$ even though the equation is meant for 1 mole.

Hence, the answer is A.