Question

An ideal gas is held in a container of volume V at pressure p. The rms speed of a gas molecule under these conditions is v. If now the volume and pressure are changed to 2V and 2p, the rms speed of a molecule will be?

a) 4v
b) v/2
c) v/4
d) 2v
e) v

Answer 1

d) 2v

Step-by-step explanation:

Since, root mean square speed of a molecule,

`V_(rms)=\sqrt{(3kT)/(m)}`

Where,

k = Boltzmann constant,

T = temperature of gas,

m = mass of molecule,

Also, the temperature of a gas,

`T=(pV)/(nR)`

Where,

p = pressure of the gas,

V = volume,

n = number of moles of gas,

R = universal gas constant,

`\implies V_(rms)=\sqrt{(3kpV)/(mnR)}`

If V = 2V and p = 2P

Then,

`V_(rms)=\sqrt{(12kpV)/(mnR)`

`=2\sqrt{(3kpV)/(mnR)`

Hence, if rms speed of a gas molecule under initial conditions is v then rms speed of a molecule will be 2v

i.e. option d is correct.