Question

For an ideal gas at 300 K and 2 atm, if the temperature is raised to 600 K at constant pressure, the volume will:

• A. Doubles
• B. Halves
• C. Remains same
• D. Quadruples

Hint:

Always ensure that temperature values are converted to absolute scale (Kelvin) before applying gas laws.
In this question, the temperatures are already given in Kelvin, allowing direct substitution.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question deals with ideal gas behaviour under changing thermodynamic conditions.
Specifically, we need to determine the effect of doubling the absolute temperature on the volume of an ideal gas while holding the pressure constant.

Step 2: Key Formula or Approach:
The state of an ideal gas is governed by the Ideal Gas Equation:
\[ P \cdot V = n \cdot R \cdot T \]
where \( P \) is pressure, \( V \) is volume, \( n \) is number of moles, \( R \) is the universal gas constant, and \( T \) is the absolute temperature (in Kelvin).
At constant pressure \( P \) and constant mass/moles \( n \), this equation reduces to Charles's Law:
\[ V \propto T \quad \implies \quad \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

Step 3: Detailed Explanation:

• Identify the given states:
Initial Temperature, \( T_1 = 300 \text{ K} \)
Final Temperature, \( T_2 = 600 \text{ K} \)
Initial Pressure, \( P_1 = 2 \text{ atm} \)
Final Pressure, \( P_2 = 2 \text{ atm} \) (constant pressure)

• Set up the ratio using Charles's Law:
\[ \frac{V_2}{V_1} = \frac{T_2}{T_1} \]

• Substitute the temperature values into the ratio:
\[ \frac{V_2}{V_1} = \frac{600 \text{ K}}{300 \text{ K}} = 2 \]

• Rearrange to find the final volume:
\[ V_2 = 2 \cdot V_1 \]


Step 4: Final Answer:
The volume of the ideal gas doubles when the absolute temperature is raised from 300 K to 600 K at constant pressure.