Question

The graph of pressure P and volume V of 1 mole of an ideal gas at constant temperature is

• A. Graph-I
• B. Graph-II
• C. Graph-III
• D. Graph-IV

Hint:

Remember the shapes of P-V diagrams for common processes:
- Isothermal: Rectangular hyperbola ($PV = \text{constant}$)
- Adiabatic: Steeper hyperbola ($PV^\gamma = \text{constant}$)
- Isobaric: Horizontal line ($P = \text{constant}$)
- Isochoric: Vertical line ($V = \text{constant}$)

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks to identify the correct Pressure-Volume (P-V) graph for an ideal gas undergoing an isothermal process (constant temperature).
Step 2: Key Formula or Approach:
For an ideal gas, the ideal gas law is $PV = nRT$.
In an isothermal process, temperature ($T$) is constant, and for 1 mole, $n=1$. So, $PV = RT = \text{constant}$.
This implies $P \propto \frac{1}{V}$. This is the mathematical form of Boyle's Law.
Step 3: Detailed Explanation:
A graph of $P$ versus $V$ where their product is constant ($P = \frac{\text{constant}}{V}$) will be a rectangular hyperbola.
Let's examine each graph:
- Graph-I: Shows pressure (P) remaining constant as volume (V) changes. This represents an isobaric process.
- Graph-II: Shows a linear relationship where $P \propto V$. This is not isothermal.
- Graph-III: Shows a linear relationship where $P$ decreases linearly with $V$. This is also not isothermal.
- Graph-IV: Shows a curve where $P$ decreases as $V$ increases, and the product $PV$ remains constant. This is characteristic of a rectangular hyperbola, which correctly represents an isothermal process.
Step 4: Final Answer:
Graph-IV correctly depicts the relationship between pressure and volume for an ideal gas at constant temperature.