Question

At \( T(K) \), the \( P, V \) and \( u_{\text{rms}} \) of 1 mole of an ideal gas were measured. The following graph is obtained. What is its slope (\( m \))? (x-axis = \( PV \); y-axis = \( u_{\text{rms}}^2 \); \( M \) = Molar mass)

• A. \( \frac{3}{M} \)
• B. \( \frac{M}{3} \)
• C. \( \left(\frac{M}{3}\right)^{1/2} \)
• D. \( \left(\frac{3}{M}\right)^{1/2} \)

Hint:

For an ideal gas, \( u_{\text{rms}}^2 = \frac{3PV}{M} \), giving a slope of \( \frac{3}{M} \) when plotted against \( PV \).

Answer 1

The Correct Option is A

Step 1: Apply Ideal Gas Law For one mole of an ideal gas: \[ PV = RT \] Step 2: Use RMS Velocity Formula The root mean square velocity is: \[ u_{\text{rms}} = \sqrt{\frac{3RT}{M}} \] Squaring both sides: \[ u_{\text{rms}}^2 = \frac{3RT}{M} \] Step 3: Express in Terms of \( PV \) Since \( PV = RT \), we substitute: \[ u_{\text{rms}}^2 = \frac{3PV}{M} \] Comparing with the straight-line equation \( y = mx \): \[ m = \frac{3}{M} \] Thus, the correct answer is \( \frac{3}{M} \).

Answer 2

Given:
- \( T \) = Temperature in Kelvin
- \( P \) = Pressure
- \( V \) = Volume
- \( u_{rms} \) = Root mean square velocity of gas molecules
- M = Molar mass of the gas

We are given a graph with:
- x-axis = \( PV \)
- y-axis = \( u_{rms}^2 \)

We need to find the slope \( m \) of the graph.

Step 1: Recall the ideal gas law for 1 mole of gas:
\[ PV = RT \]

Step 2: The root mean square velocity of an ideal gas is given by:
\[ u_{rms} = \sqrt{\frac{3RT}{M}} \] where \( M \) is the molar mass.
Squaring both sides,
\[ u_{rms}^2 = \frac{3RT}{M} \]

Step 3: Express \( u_{rms}^2 \) in terms of \( PV \):
Since \( PV = RT \), substitute \( RT = PV \) in the equation:
\[ u_{rms}^2 = \frac{3PV}{M} \]

Step 4: Writing the equation of the straight line (graph):
\[ y = m x \] where \( y = u_{rms}^2 \) and \( x = PV \), so comparing:
\[ m = \frac{3}{M} \]

Therefore, the slope \( m \) of the graph is:
\[ \boxed{\frac{3}{M}} \]