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^oC to -66 ^oC.

The number of moles must be the same throughout this change. Also, the volume hasn't changed 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 v_avg is given by square root of (8RT/M) where M is the molar mass , R = gas constant and T is the temperature of the gas. So, KE_avg = 4RT/

So, KE_avg/T = constant

Initial KE_avg/ initial T = final KE_avg / final T

So, KE_i/T_i = KE_f/T_f we get KE_f = 0.639 K_i

u_rms is square root of 3RT/M So, u_rms/T1/2 = constant.

u_i/T_i1/2 = u_f/T_f2 so we get u_f = 0.799u_i

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 KE_avg and u_rms depends on the 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.