The RMS velocity (\(v_{{rms}}\)) of one mole of an ideal gas was measured at different temperatures and the following graph is obtained. What is the slope (m) of the straight line?
Show Hint
The relationship between the RMS velocity and temperature for an ideal gas is given by the equation \( v_{{rms}} = \sqrt{\frac{3RT}{M}} \). The slope of the straight line in a temperature versus RMS velocity graph will be proportional to \( \frac{3R}{M} \).
The root-mean-square (RMS) velocity of an ideal gas is given by the equation:
\( v_{rms} = \sqrt{\frac{3RT}{M}} \)
Here, \( R \) is the universal gas constant, \( T \) is the absolute temperature, and \( M \) is the molar mass of the gas. To find the relationship between RMS velocity and temperature, we square both sides of the equation:
\( v_{rms}^2 = \frac{3RT}{M} \)
Rearranging, we get:
\( T = \frac{M}{3R} v_{rms}^2 \)
This resembles the equation of a straight line \( y = mx + c \), where \( T \) is on the y-axis and \( v_{rms}^2 \) is on the x-axis. Comparing both equations:
\( y \equiv T \)
\( x \equiv v_{rms}^2 \)
\( m = \frac{M}{3R} \)
\( c = 0 \) (since no additional terms are present)
In the provided graph, since \( v_{rms}^2 \) is plotted against \( T \), the slope \( m \) of the straight line represents:
\( m = \frac{M}{3R} \)
Therefore, the solution states that the slope cannot be \(\frac{M}{3R}\) as given in some options. To find the correct correspondence, the correct slope \( m \) should reflect:
\( m = \frac{3R}{M} \)
Thus, the option showing \( \frac{3R}{M} \) is indeed the correct answer, representing the direct relationship of temperature with RMS velocity squared.
Was this answer helpful?
0
0
Show Solution
Verified By Collegedunia
Approach Solution -2
We are given the RMS velocity (\(v_{{rms}}\)) for one mole of an ideal gas at different temperatures. From the graph, we can see that the relationship between the temperature (\(T\)) and the RMS velocity is linear. The equation for RMS velocity for an ideal gas is given by:
\[
v_{{rms}} = \sqrt{\frac{3RT}{M}}
\]
where:
- \( R \) is the universal gas constant,
- \( M \) is the molar mass of the gas,
- \( T \) is the temperature.
The equation of the line in the graph is of the form \( y = mx + c \), where \( m \) is the slope. In this case, the temperature \( T \) is plotted on the x-axis and the RMS velocity \( v_{{rms}} \) is plotted on the y-axis. Since the equation \( v_{{rms}} = \sqrt{\frac{3RT}{M}} \) is a linear equation, the slope \( m \) is the derivative of \( v_{{rms}} \) with respect to \( T \).
Taking the derivative with respect to \( T \) gives:
\[
\frac{d(v_{{rms}})}{dT} = \frac{d}{dT} \left( \sqrt{\frac{3RT}{M}} \right) = \frac{1}{2} \cdot \left( \frac{3R}{M} \right)^{1/2}
\]
Thus, the slope \( m \) is:
\[
m = \frac{3R}{M}
\]
Thus, the correct slope of the straight line is \( \frac{3R}{M} \).
