Question

An ideal gas molecule is present at $27^\circ$ C. By how many degree centigrade its temperature should be raised so that its ${V_{rms}}$ , ${V_{mp}}$ and ${V_{av}}$ all may double?
(A) $900^\circ$ C
(B) $108^\circ$ C
(C) $927^\circ$ C
(D) $81^\circ$ C

Answer 1

Hint : Basically, ideal gas is a gas which consists of particles that are randomly moving and have no interparticle interactions. Now, here in this only the temperature of the ideal gas molecule is given, so we will use a formula to solve this by manipulating that.

Complete Step By Step Answer:
Now, as we can see that all the options are given to us in Celsius, so first we are going to convert Celsius to kelvin. For that we have a relation, that is, $K = C + 273$
So, using this formula we will convert the temperature from the Celsius to kelvin:
Given temperature is $27^\circ C$ , so-
$K = 27 + 273 = 300K$
Now, we will see that how ${V_{rms}}$ , ${V_{mp}}$ and ${V_{av}}$ are related to temperature:
Root mean square velocity ( ${V_{rms}}$ ) = $\sqrt {\dfrac{{3RT}}{M}}$ , here we can see that ${V_{rms}}$ $\alpha \sqrt T$
Most probable velocity ( ${V_{mp}}$ ) = $\sqrt {\dfrac{{2RT}}{M}}$ , here ${V_{mp}}$ $\alpha \sqrt T$
Average velocity ( ${V_{av}}$ ) $= \sqrt {\dfrac{{8RT}}{{\pi M}}}$ , here also ${V_{av}}\alpha \sqrt T$
We can see from these relations that if we will increase the velocity two times, that is, if we double the velocity then the temperature will increase four times.
$\therefore$ $T' = 4T$
$T' = 4 \times 300$ $= 1200K$
Now, we will convert the result into Celsius again using –
$C = K - 273$
$C = 1200 - 273 = 927^\circ C$
So, the correct option is option C.

Note :
Here in this we need to convert the temperature into kelvin first and then we will put those values in the formula to get the answer. These velocities help us to understand the rate at which the molecules are moving in the particular temperature.