A sealed bottle at 1 atm of pressure holds 1 mole of neon and 1 mole of argon ga

Question

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.

A.) Find the number of molecules of neon which have velocities between 750 and 1000 (m/s).

*******Answer is 9.613 x 10^22

B.) Find the volume of the bottle.

C.) In the bottle, vrms of the neon molecules is 603.9 m/s. The temperature is increased by 29.5 deg C, find the new v_rms of these molecules.

*******Answer is 6.333 x 10^2 m/s

D.) Find the new pressure in the bottle after this increase in temperature (in atm).

I have figured out parts A and C, but I cannot figure out parts B and D! Thanks!

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 number of molecules of argon which have velocities between 750 and 1000 (m/s). *******Answer is 9.613 Times 10^22 Find the volume of the bottle. In the bottle, vrms of the neon molecules is 603.9 m/s. The temperature is increased by 29.5 deg C, find the new v_rms of these molecules. *******Answer is 6.333 Times 10^2 m/s Find the new pressure in the bottle after this increase in temperature (in atm).

Explanation / Answer

B) by ideal gas equation

PV=nRT

Given P=1 atm

T=295 K

moles n = moles of argon + moles of neon

Given moles of argon =1

moles of neon = 1

n= 1+1

n=2

R=0.0821

V=nRT/P

V= 2 x 0.0821 x 295 /1

V= 48.44 L

So the volume is 48.44 L

D) Given the temp is raised by 29.5 C

final temp T2 = 295 + 29.5

T2= 324.5

The volume is constant

Using charles second law

so P1/T1 =P2/T2

P2= P1T2/T1

P2= 1 x 324.5 / 295

P2=1.1

So final pressure is 1.1 atm

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