Question

A sample of 4.63 mol of xenon is confined at low pressure in a volume at a temperature of 51 °C. Describe quantitatively the effects of each of the following changes on the pressure, the average kinetic energy per molecule in the gas, and the root-mean-square speed.

(a) The temperature is decreased to -66 °C.
(b) The volume is halved.
(c) The amount of xenon is increased to 6.02 mol.

Give each answer as a decimal factor of the form: new value = factor old value.   A factor of 1 means no change.

Change	P	KEavg	urms
(a)
(b)
(c)

Answer 1

In the first change we have reduced the temperature of the gas from 51 degree C to -66 degree C.

The number of moles must be the same throughout this change. Also, the volume hasn't change too.

So, we can use the relation that the ratio of P and T must be constant as n,V= constant.

So, P1/T1 = P2/T2 , T1 = 324K, T2 = 207K P1 = initial pressure and P2 = final pressure.

P2 = 0.639P1 so, the pressure has reduced.

KE_avg = mv_avg²/2 where m is the mass of the gas and vavg is given by square root of (8RT/M) where M is the molar mass , R = gas constant and T is temperature of the gas. So, KEavg = 4RT/

So, KEavg/T = constant

Initial KEavg/ initial T = final KEavg / final T

So, KEi/Ti = KEf/Tf we get KEf = 0.639 Ki

u_rms is sqaure root of 3RT/M So, urms/T^1/2 = constant.

ui/Ti^1/2 = uf/Tf² so we get uf = 0.799ui

b)

By PV=nRT if volume is halved the pressure will increase two times and the temperature will remain the same by Boyle's law.

So, Pf = 2Pi

As KEavg and urms depends on temperature they will remain the same as temperature is constant.

c)

As we increase the number of moles of gas the pressure will increase for an instant and then decrease with the expansion of the container walls such that V/n = constant by Avogadro's law.

So, after sometime the pressure will again comeback to the initial state. Also, the temperature will remain the same. So, no effect on each quantity given.