Question

The ratio of rms speed and most probable speed of molecules of an ideal gas is R at temperature T. If the temperature is increased by 2% then the percentage change in the ratio will be

Answer 1

Given:

The ratio of rms speed and most probable speed of molecules of an ideal gas is R at temperature T. If the temperature is increased by 2%

To find:

The percentage change in the ratio

Solution:

To solve the given problem we will use the following formula of the root mean square speed and the most probable speed of a gas:

$\boxed{\bold{V_r_m_s = \sqrt{\frac{3RT}{M} } }}\\\\\boxed{\bold{V_m_p_s = \sqrt{\frac{2RT}{M} } }}$  

where R = gas constant = 8.314 kg m²/mol K s², T = temperature (K), M = molecular mass (kg/mol)

The ratio of root mean square speed and most probable speed of molecules of an ideal gas is R at temperature T

i.e., $\frac{\sqrt{\frac{3RT}{M}} }{\sqrt{ \frac{2RT}{M} }} = R$

$\implies \frac{\sqrt{3} \times \sqrt{\frac{RT}{M}} }{\sqrt{2}\times \sqrt{ \frac{RT}{M} }} = R$

$\implies R = \frac{\sqrt{3} }{\sqrt{2}}$ . . . . (1)

If the temperature is increased by 2%

Then the increased temperature, T' = $T + \frac{2T}{100} = 1.02T$

Therefore,

The new ratio (R') will be,

= $\frac{\sqrt{\frac{3RT'}{M}} }{\sqrt{ \frac{2RT'}{M} }}$

= $\frac{\sqrt{\frac{3R(1.02T)}{M}} }{\sqrt{ \frac{2R(1.02T)}{M} }}$

= $\frac{\sqrt{3} \times \sqrt{\frac{R(1.02T)}{M}} }{\sqrt{2}\times \sqrt{ \frac{R(1.02T)}{M} }}$

= $\frac{\sqrt{3} }{\sqrt{2}}}$

from (1), we get

= R  . . . (2)

Now,

The percentage change in the ratio will be,

= $\frac{New \:Ratio \:-\: Old\:Ratio}{Old\:Ratio} \times 100$

= $\frac{R' - R}{R} \times 100$

from (2), we get

= $\frac{R- R}{R} \times 100$

= $\bold{0}$ ← option (4)

Thus, the percentage change in the ratio will be → Zero.

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