Question

The rms velocity of gas molecules is increased from V to 3V, when the temperature of gas molecules increases
from x°c to87°c. Find the value of x.​

Answer 1

Explanation:

We know that the rms velocity is given as V

rms

=

2

M

3RT

We also know that increasing temperature will increase pressure so we just need to see the formula for temperature dependence and just apply above formula.

From above formula

v

2

v

1

=

2

T

2

T

1

=

2

127+273

27+273

=

2

4

3

so v

2

=

2

3

2v

1

=

3

400

m/s

Answer 2

The value of x is  - 233 °C.

Given: Initial RMS velocity of the gas(u) = V

The final RMS velocity of the gas (v) = 3V

Initial temperature  = x°C

Final temperature = 87°C

To Find: the value of x

Solution:

The  velocity of a gas is given as

v = $\sqrt{\frac{3RT}{M} }$, where v is the RMS velocity, R is the gas constant, T is the temperature(in Kelvin) and m is the molecular mass of the gas.

Temperature in Kelvin = temperature in Celcius + 273

Initial Temperature in Kelvin(T1) = (x +273)K

Final temperature in Kelvin (T2) = 87 + 273

                                                     = 360K

Since the gas is the same so the molecular mass will be the same

u = $\sqrt{\frac{3R(T1)}{M} }$                      

V =  $\sqrt{\frac{3R(T1)}{M} }$                      ..1

v = $\sqrt{\frac{3R(360)}{M} }$

3V =  $\sqrt{\frac{3R(360)}{M} }$                   ..2

Dividing equation 1 by equation 2, we get

$\frac{1}{3} = \sqrt{\frac{T1}{360} }$

On squaring both sides we get

1/ 9 = T1 / 360

T1 = 360 / 9

T1 = 40K

T1 = x + 273

40 = x + 273

x  = 40 - 273

x = - 233

Therefore, the value of x is  - 233 °C.