Q.Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity $25.0\ \text{m}^3$ at a temperature of $27\ ^\circ\text{C}$ and $1\ \text{atm}$ pressure.

Concept & Intuition

Standard Temperature and Pressure (STP) is a fixed reference state: 0 °C and 1 atm. A cornerstone of ideal‑gas behaviour is that one mole of any gas occupies 22.4 L at STP. This is independent of the gas’s identity — a mole of hydrogen and a mole of oxygen each take up the same volume.

Here we have a mixture of two gases in a known total volume. If we can find the total number of moles in the vessel, and we already know the number of moles of hydrogen from its mass, then the difference gives the moles of oxygen. Converting those moles back to mass yields the answer.

1. Convert the vessel’s volume to litres The volume is given as $27 \times 10^4 \ \mathrm{cc}$. Since $1 \ \mathrm{L} = 1000 \ \mathrm{cc}$,

2. Find the total number of moles of gas in the vessel At STP, 1 mol occupies 22.4 L. Therefore,

Compute:

3. Find the number of moles of hydrogen Mass of hydrogen = 16 g, molar mass = 2 g mol⁻¹.

4. Determine the moles of oxygen The mixture contains only hydrogen and oxygen, so

5. Convert moles of oxygen to mass Molar mass of oxygen ($\mathrm{O_2}$) = 32 g mol⁻¹. …

Concept & Intuition

The problem gives equal masses of oxygen (O₂) and hydrogen (H₂) at the same temperature, but different volumes. Pressure depends on the number of gas molecules (moles), not the mass directly. Since oxygen has a much larger molar mass than hydrogen, the same mass of oxygen contains far fewer moles. Combined with the volume difference, this drastically affects the pressure ratio. The ideal gas law $PV = nRT$ is the perfect tool here.

Step-by-step solution

1. Write the ideal gas law for each vessel.

   For vessel A (oxygen): $P_A V_A = n_A R T$

   For vessel B (hydrogen): $P_B V_B = n_B R T$

   Temperature $T$ and gas constant $R$ are the same for both.

2. Relate the number of moles to the given masses.

   Let the common mass be $m$.

   Molar mass of O₂ = 32 g/mol, of H₂ = 2 g/mol.

   So

3. Use the volume relation.

   Given $V_B = 2 V_A$. Let $V_A = V$, then $V_B = 2V$.

4. Find the pressure ratio.

   From the ideal gas law:

Take the ratio:

$\frac{P_A}{P_B} = \frac{n_A R T / V}{n_B R T / (2V)} = \frac{n_A}{n_B} \cdot 2.$ …

Step 1 — Apply the ideal gas law at constant T

For $n=1$ mole of ideal gas held at constant temperature $T$ (an isothermal process):

Step 2 — Identify the shape of the curve

An inverse relationship $P \propto 1/V$ is a rectangular hyperbola: $P$ starts high at small $V$ and decreases smoothly and non-linearly as $V$ increases, never touching either axis. This matches Graph-IV, described as a curve starting high on the $P$-axis and decreasing smoothly toward the $V$-axis. …

The coefficient of volume expansion, usually denoted $\gamma$ or $\beta$, is defined as the fractional change in volume per unit change in temperature at constant pressure:

But here the gas is not expanding at constant pressure — instead, pressure and temperature are linked by $P T^2 = \text{constant}$. So we cannot simply use the ideal gas law with $P$ fixed. Instead, we must find how $V$ depends on $T$ along the given path, then differentiate.

1. Write the given condition and the ideal gas law.

The ideal gas law:

Given: $P T^2 = k$ (a constant). So $P = \frac{k}{T^2}$.

2. Substitute $P$ into the ideal gas law to eliminate $P$.

Multiply both sides by $T^2$:

So

This shows that along this expansion, volume is proportional to $T^3$.

3. Differentiate $V$ with respect to $T$.

But from step 2, $\frac{nR}{k} = \frac{V}{T^3}$. Substitute:

4. Compute the coefficient of volume expansion.

Convert the data to SI units:

• $V = 300\ \text{cm}^3 = 3.0\times10^{-4}\ \text{m}^3$
• $P = 1\ \text{mm Hg} = 133.3\ \text{Pa}$
• $T = 27^\circ\text{C} = 300\ \text{K}$

Number of moles: …

Concept: the coefficient of volume expansion is defined as

gamma = (1/V) * (dV/dT) at constant pressure.

For an ideal gas at constant pressure, PV = nRT => V = nRT/P, so

(dV/dT) at constant P = nR/P = V/T.

Hence gamma = (1/V) * (V/T) = 1/T. …