Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

SUUUPPEERR HARD. I have no idea how to do this one. Please include work for all

ID: 2242861 • Letter: S

Question



SUUUPPEERR HARD. I have no idea how to do this one. Please include work for all credit and points. WILL RATE!




A sealed bottle at 1 atm of pressure holds 1 mole of neon and 1 mole of argon gas at a temperature of 295 K. The curves show the distributions of velocities of the molecules of each gas. Find the molecules of neon which have velocities between 750 and 1000 (m/s). Find the volume of the bottle. In the bottle, vrms of the neon molecules is 603.9 m/s. The temperature is increased by 20.4o deg C, find the new vrms of these molecules. Find the new pressure in the bottle after this increase in temperature (in atm).

Explanation / Answer

A. A= 1/2 (b1+b2) h

A= 1/2 (b1+b2) (1000 m/s -750 m/s) I cannot see what the curve shows, but add the two numbers.



B. PV=nRT

P= 1 atm

n= # moles of neon + # of moles argon, which is 2

R= 0.0821

T= 295K to Celsius is 21.85 C.

21.85 C + 20.4 C = 42.25 C

42.25 C to Kelvin is 315.4 K


V= (2*0.0821*315.4K) / 1 atm = 51.78868m^3



C. Use vrms = (3RT/M)^(1/2)

R= constant 8.3145

M= molecular mass of Neon (20.1797)

(3*8.3145* 315.4K) / (20.1797 / 1000) ^(0.5)

== 6.244 x 10^2 m/s







D. PV=nRT


P * 51.79 = 2 * 0.082 * 315.4K


P= 0.998756 atm or 1.00 atm