Question

At \( T(K) \), the \( P, V \) and \( u_{{rms}} \) of 1 mole of an ideal gas were measured. The following graph is obtained. What is its slope (\( m \))? (x-axis = \( PV \); y-axis = \( u_{{rms}}^2 \); \( M \) = Molar mass) \includegraphics[]{126.png}

• 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, \( u_{{rms}}^2 = \frac{3PV}{M} \), 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: \[ u_{{rms}} = \sqrt{\frac{3RT}{M}} \] Squaring both sides: \[ u_{{rms}}^2 = \frac{3RT}{M} \] Step 3: Express in Terms of \( PV \) Since \( PV = RT \), we substitute: \[ u_{{rms}}^2 = \frac{3PV}{M} \] Comparing with the straight-line equation \( y = mx \): \[ m = \frac{3}{M} \] Thus, the correct answer is \( \frac{3}{M} \).