Question

At \( T(K) \) for one mole of an ideal gas, the graph of \( P \) (on y-axis) and \( V^{-1} \) (on x-axis) gave a straight line with slope of \( 32.8 \) L atm mol\(^{-1} \). What is the temperature (in K)? \((R = 0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1})\)

• A. \( 600 \)
• B. \( 200 \)
• C. \( 800 \)
• D. \( 400 \)

Hint:

For an ideal gas, when plotting \( P \) versus \( V^{-1} \), the slope gives \( nRT \). This can be used to determine temperature if the slope and \( R \) are known.

Answer 1

The Correct Option is D

Step 1: Understanding the Graph Relationship
For an ideal gas, the equation is: \[ PV = nRT \] Rearranging in the form of a straight line equation:
\[ P = \frac{nRT}{V} \] which can be rewritten as: \[ P = (nR T) \times V^{-1} \] Comparing with the equation of a straight line \( y = mx + c \), we see that the slope of the graph is equal to \( nRT \).
Step 2: Calculating Temperature
Given that: - Slope \( = 32.8 \) L atm mol\(^{-1}\)
- \( n = 1 \) (one mole of gas)
- \( R = 0.0821 \) L atm mol\(^{-1}\) K\(^{-1}\)
Using: \[ \text{Slope} = nRT \] \[ 32.8 = (1) \times (0.0821) \times T \] Solving for \( T \): \[ T = \frac{32.8}{0.0821} \] \[ T = 400 \text{ K} \] Final Answer: The correct temperature is \( 400 \) K, which matches Option (4).