Question

a sealed cylinder contains a sample of ideal gas at a pressure of 2.0 atm. The rms speed of the molecules is v0. If the rms speed is then reduced to 0.90 v0, what is the pressure of the gas?

Answer 1

Answer:

P_2 = 1.62 atm

Explanation:

We know the formula for the rms speed of the ideal gas is  given by

$v_{rsm}=\sqrt{\frac{3PV}{m} }$

P= pressure of the surrounding

V= volume of the vessel

m= mass of the gas

Now, From this formula rms speed (v_rms) is directly proportional to square root is pressure.

Then  

$\frac{v_{rsm,1}}{v_{rsm,2}} =\sqrt{\frac{P_1}{P_2} }$

given that v_rsm,1= v0

and v_rsm,2=0.9v0

putting these values we get

$\frac{v0}{0.9v0} =\sqrt{\frac{2}{P_2} }$

P_2 = 1.62 atm

Answer 2

Answer:

The pressure of the gas is 1.8 atm.

Explanation:

Given that,

Pressure of ideal gas= 2.0 atm

rms speed of the molecule = v₀

Reduced rms speed = 0.90 v₀

We know the formula of rms speed of the ideal gas

$v_{rms}=\sqrt{\dfrac{3Pv}{m}}$

From this formula rms speed is directly proportional to square root is pressure.

We need to calculate the pressure of the gas

Using formula of rms speed

$\dfrac{v_{rms}}{v_{rms}}=\sqrt{\dfrac{P_{1}}{P_{2}}}$

Put the value into the formula

$\dfrac{v_{0}}{0.90v_{0}}=\sqrt{\dfrac{2.0}{P_{2}}}$

$P_{2}=2.0\times0.90$

$P_{2}=1.8\ atm$

Hence, The pressure of the gas is 1.8 atm.