Question

Q 42. Ratio of temperature and nelecular mass (T./Mp) of gas P is 9 times the ratio of temperature and molecular mass (TQM) of gas Q. The
ratio of RMS speed of gas Q to that of gas P is:
Ops:
A. 01:3
B. 04:3
c. 03:4
D. 03:1
has more than two decimal

Answer 1

Answer:

03:1h few of the united states in which t to the united nations in the th and

Answer 2

Answer:

The ratio of root mean square (RMS) speed of gas Q to that of gas P is 3 : 1.

Therefore, option (D) is correct.

Explanation:

Given, the ratio of temperature and molecular mass of gas P is 9 times the ratio of temperature and molecular mass of gas Q.

Consider that, $T_P$  and $M_P$ are the temperature and molecular mass of gas P respectively. The  $T_Q$  and $M_Q$ are the temperature and molecular mass of gas Q respectively.

Therefore, $\frac{T_P}{M_P} =9\times\frac{T_Q}{M_Q}$                                                       ..................(1)

The root mean square (RMS) velocity of a gas can be calculated by:

$V_{rms}=\sqrt{\frac{3RT}{M}}$

$\frac{(V_{rms})_P}{(V_{rms})_Q} =\frac{\sqrt{\frac{3RT_P}{M_P}}}{\sqrt{\frac{3RT_Q}{M_Q}}}$

$\frac{(V_{rms})_P}{(V_{rms})_Q} =\frac{\sqrt{\frac{T_P}{M_P}}}{\sqrt{\frac{T_Q}{M_Q}}}$

Substitute the value ration of temperature and molecular mass from equation (1):

$\frac{(V_{rms})_P}{(V_{rms})_Q} =\frac{\sqrt{9\times\frac{ T_Q}{M_Q}}}{\sqrt{\frac{T_Q}{M_Q}}}$

$\frac{(V_{rms})_P}{(V_{rms})_Q} =\frac{\sqrt{9} }{1}$

$\frac{(V_{rms})_P}{(V_{rms})_Q} =\frac{3 }{1}$

Therefore, the ratio of RMS speed of gas Q to that of gas P is 3 : 1.

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