Question

The variation of volume of an ideal gas with its number of moles (\( n \)) is obtained as a graph at 300 K and 1 atm pressure. What is the slope of the graph? 

• A. \( 24.6 \) L
• B. \( 24.6 \) L mol\(^{-1}\)
• C. \( \frac{1}{24.6} \) L
• D. \( \frac{1}{24.6} \) L\(^{-1}\) mol

Hint:

For an ideal gas at STP (1 atm, 273 K), the molar volume is 22.4 L. At 300 K, it increases to approximately 24.6 L per mole. The slope of a volume vs. moles graph is given by \( \frac{RT}{P} \).

Answer 1

The Correct Option is B

Step 1: Use the Ideal Gas Equation 
The ideal gas equation is: \[ PV = nRT \] where: - \( P \) is the pressure (1 atm), - \( V \) is the volume, - \( n \) is the number of moles, - \( R \) is the universal gas constant (\( 0.0821 \) L atm mol\(^{-1}\) K\(^{-1}\)), - \( T \) is the temperature (300 K). 

Step 2: Express Volume as a Function of \( n \) 
Rearranging the equation: \[ V = \frac{nRT}{P} \] Since \( R = 0.0821 \), \( T = 300 \) K, and \( P = 1 \) atm: \[ V = \frac{(n)(0.0821)(300)}{1} \] \[ V = 24.6 n \] 

Step 3: Identify the Slope 
The equation is in the form: \[ V = m n \] where \( m \) (the slope) is 24.6 L mol\(^{-1}\). 

Step 4: Verify the Correct Answer 
Thus, the slope of the graph is \( 24.6 \) L mol\(^{-1}\), which matches Option (2).