State Boyle's Law and Charle's Law. Derive the relation Pv = nRT . An evacuated glass vessel weighs 50 gm when empty, 148 gm when filled with a liquid of density 0.98 gm/ml, and 50.5 gm when filled…

Model Answer

1. Boyle’s Law and Charles’s Law

Boyle’s Law

Boyle’s Law states that for a fixed amount of an ideal gas at constant temperature, the pressure (P) of the gas is inversely proportional to its volume (V). Mathematically, it is expressed as:

$$P\propto \frac{1}{V}\text{(at constant T and n)}$$

Or,

$$P_1{V}_1=P_2{V}_2\text{(for a fixed amount of gas at constant temperature)}$$

Explanation:

• If the volume of a gas decreases, its pressure increases, provided the temperature and amount of gas remain constant.
• This law is derived from the kinetic theory of gases, where reduced volume leads to more frequent collisions of gas molecules with the container walls, increasing pressure.

Charles’s Law

Charles’s Law states that for a fixed amount of an ideal gas at constant pressure, the volume (V) of the gas is directly proportional to its absolute temperature (T). Mathematically:

$$V\propto T\text{(at constant P and n)}$$

Or,

$$\frac{V_1}{T_1}=\frac{V_2}{T_2}\text{(for a fixed amount of gas at constant pressure)}$$

Explanation:

• As temperature increases, the kinetic energy of gas molecules increases, causing them to occupy more space (increase in volume) if pressure is constant.
• This law is valid only for absolute temperatures (in Kelvin).

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2. Derivation of the Ideal Gas Equation $PV=nRT$

The ideal gas equation combines Boyle’s Law, Charles’s Law, and Avogadro’s Law. The derivation is as follows:

Step 1: Boyle’s Law (Pressure-Volume Relationship)

From Boyle’s Law: $$P\propto \frac{1}{V}\Rightarrow P=\frac{k_1}{V}$$ where $k_1$ is a proportionality constant.

Step 2: Charles’s Law (Volume-Temperature Relationship)

From Charles’s Law: $$V\propto T\Rightarrow V=k_{2}T$$ where $k_2$ is another proportionality constant.

Step 3: Combining Boyle’s and Charles’s Laws

Substituting $V=k_{2}T$ into Boyle’s equation: $$P=\frac{k_1}{k_{2}T}=\frac{k}{T}$$ where $k=\frac{k_1}{k_2}$ is a new constant.

But this is incomplete because it does not account for the amount of gas (n). Avogadro’s Law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Thus, volume is also proportional to the number of moles ($n$):

$$V\propto n\Rightarrow V=k_{3}n$$

Step 4: Final Combination

Now, combining all three relationships: $$PV=nRT$$ where:

• $P$ = Pressure (atm or mmHg)
• $V$ = Volume (L)
• $n$ = Number of moles (mol)
• $R$ = Universal gas constant ($0.0821{\text{L atm K}}^{-1}{\text{mol}}^{-1}$)
• $T$ = Absolute temperature (K)

Explanation of Terms:

• $R$ is a constant that depends on the units used (e.g., $0.0821{\text{L atm K}}^{-1}{\text{mol}}^{-1}$ or $8.314{\text{J mol}}^{-1}{\text{K}}^{-1}$).
• The equation $PV=nRT$ is the Ideal Gas Law, which describes the behavior of an ideal gas under any conditions of $P,V,T,$ and $n$.

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3. Determination of Molecular Mass of the Gas

Given Data:

• Mass of empty vessel = 50 g
• Mass of vessel + liquid = 148 g
• Density of liquid = 0.98 g/mL
• Mass of vessel + gas = 50.5 g
• Pressure of gas ($P$) = 760 mmHg = 1 atm
• Temperature ($T$) = 27°C = 300 K (since $273+27=300$)

Step 1: Calculate the Volume of the Vessel (Using Liquid Data)

The mass of the liquid alone is: $$\text{Mass of liquid}=148\text{g}-50\text{g}=98\text{g}$$

Given the density ($\rho$) of the liquid is 0.98 g/mL, the volume ($V$) of the vessel is: $$V=\frac{\text{Mass}}{\text{Density}}=\frac{98\text{g}}{0.98\text{g/mL}}=100\text{mL}=0.1\text{L}$$

Step 2: Calculate the Mass of the Gas Alone

The mass of the gas is: $$\text{Mass of gas}=50.5\text{g}-50\text{g}=0.5\text{g}$$

Step 3: Use the Ideal Gas Law to Find Moles of Gas ($n$)

The ideal gas equation is: $$PV=nRT$$

Rearranging to solve for $n$: $$n=\frac{PV}{RT}$$

Substitute the known values:

• $P=1\text{atm}$
• $V=0.1\text{L}$
• $R=0.0821{\text{L atm K}}^{-1}{\text{mol}}^{-1}$
• $T=300\text{K}$

$$n=\frac{(1)(0.1)}{(0.0821)(300)}=\frac{0.1}{24.63}\approx 0.00406\text{mol}$$

Step 4: Calculate the Molecular Mass ($M$) of the Gas

The molecular mass is given by: $$M=\frac{\text{Mass of gas (g)}}{\text{Moles of gas (mol)}}=\frac{0.5\text{g}}{0.00406\text{mol}}\approx 123.15\text{g/mol}$$

Molecular mass of the gas ≈ 123 g/mol

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4. Number of Molecules in 12.3 g of the Gas

Step 1: Calculate Moles in 12.3 g of Gas

Using the molecular mass ($M=123\text{g/mol}$): $$n=\frac{\text{Mass}}{M}=\frac{12.3\text{g}}{123\text{g/mol}}=0.1\text{mol}$$

Step 2: Calculate Number of Molecules Using Avogadro’s Number

Avogadro’s number ($N_A$) is $6.022\times 10^{23}\text{molecules/mol}$.

$$\text{Number of molecules}=n\times N_A=0.1\text{mol}\times 6.022\times 10^{23}\text{molecules/mol}$$

$$=6.022\times 10^{22}\text{molecules}$$

Number of molecules in 12.3 g of gas = $6.022\times 10^{22}$ molecules