State Boyle's Law and draw i) P vs V , ii) V vs 1p and iii) Pv vs p graphs to support Boyle's law. What is the practical significance of this law? A toy balloon blown up at 10^ , has a volume 960 cc…

Model Answer

Boyle’s Law

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

$$P\propto \frac{1}{V}\text{or}{P}_1{V}_1=P_2{V}_2$$

This law holds true only when:

• The amount of gas (number of moles) remains constant.
• The temperature is kept constant (isothermal conditions).

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Graphical Representation of Boyle’s Law

(i) $P$ vs $V$ Graph

For a fixed mass of gas at constant temperature, as the pressure increases, the volume decreases, and vice versa. The graph is a hyperbola (inverse relationship).

(ii) $V$ vs $\frac{1}{P}$ Graph

When volume is plotted against the reciprocal of pressure, the graph is a straight line passing through the origin, confirming the inverse proportionality.

(iii) $PV$ vs $P$ Graph

When the product of pressure and volume ($PV$) is plotted against pressure, the graph is a horizontal straight line (since $PV=\text{constant}$).

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Practical Significance of Boyle’s Law

Boyle’s Law has several real-world applications:

1. Respiratory System: During inhalation, the diaphragm contracts, increasing lung volume and decreasing pressure, allowing air to flow in.
2. Scuba Diving: Divers must account for pressure changes to avoid lung over-expansion injuries.
3. Automobile Airbags: Airbags use compressed gas that expands rapidly when released, following Boyle’s Law.
4. Syringe Operation: Medical syringes rely on Boyle’s Law to draw and inject fluids.
5. Weather Forecasting: Changes in atmospheric pressure affect weather patterns.

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Numerical Problem: Balloon Bursting at Different Temperatures

Given:

• Initial temperature, $T_1=10^{\circ }\text{C}=283\text{ K}$
• Initial volume, $V_1=960\text{ cc}$
• Balloon is distended to $\frac{7}{8}$ of its maximum capacity.
• Final temperature, $T_2=25^{\circ }\text{C}=298\text{ K}$

Assumptions:

• The balloon behaves like an ideal gas.
• The maximum volume ($V_{\text{max}}$) is when the balloon is fully stretched.
• The balloon bursts when it reaches $V_{\text{max}}$.

Step 1: Determine Maximum Volume

Since the balloon is at $\frac{7}{8}$ of its maximum capacity: $$V_1=\frac{7}{8}{V}_{\text{max}}⟹V_{\text{max}}=\frac{8}{7}\times 960=1085.71\text{ cc}$$

Step 2: Check if Balloon Bursts at $25^{\circ }\text{C}$

Using Charles’s Law (since pressure is constant in this scenario): $$\frac{V_1}{T_1}=\frac{V_2}{T_2}$$ $$V_2=V_1\times \frac{T_2}{T_1}=960\times \frac{298}{283}=1023.32\text{ cc}$$ Since $1023.32\text{ cc}<1085.71\text{ cc}$, the balloon does not burst at $25^{\circ }\text{C}$.

Step 3: Calculate Temperature at Which Balloon Bursts

Let $T$ be the temperature at which the balloon bursts (reaches $V_{\text{max}}$): $$\frac{V_1}{T_1}=\frac{V_{\text{max}}}{T}$$ $$T=T_1\times \frac{V_{\text{max}}}{V_1}=283\times \frac{1085.71}{960}=323.15\text{ K}=50^{\circ }\text{C}$$

Final Answer:

The balloon will not burst at $25^{\circ }\text{C}$. It will burst when the temperature reaches $50^{\circ }\text{C}$.