A sealed bottle at 1 atm of pressure holds 1 mole of neon gas and 1 mole of argo

Question

A sealed bottle at 1 atm of pressure holds 1 mole of neon gas and 1 mole of argon gas at a temperature of 295 K. The curves in the Figure show how the speeds are distributed for the molecules of each gas. Find the approximate number of molecules of neon that have velocity between 200 m/s and 400 m/s. Find the volume of the bottle. The root-mean-square velocity of the neon molecules in the bottle is 603.9m/s at 295K. The temperature is then increased by 10.0C?. Find the new vrmsof these molecules. What is the pressure in the bottle after this increase in temperature? Select E -Equal to, G -Greater than, L -Less than. If the first is G and the rest are L enter GLLLLLL. A) The rate of collisions of neon against the bottle is ___ the rate of collisions of argon. B) vrmsof argon is ___ vrmsof neon. C) The pressure of argon is ___ the pressure of neon. D) avg KE of neon is ___ avg KE of argon. E) vrmsof neon is ___ most probable v of neon. F) most probable v of neon is ___ most probable v of argon. G) The total mass of neon gas is ___ the total mass of argon gas.

Explanation / Answer

Neon has a lower atomic number than argon, so we can assume it have its peak at a higher number. This means we are interested in the shorter peak.

To find the number of particles we need to integrate, or get the area, under this shorter curve from 200 to 400. We can assume this section of the curve to be a trapezoid with one side length of 4 and the other to 9. so now we can get the area by:

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

h is the difference in the x axis and we already know the base lengths. so we plug them in.

A=(1/2)(4+9)x1020(400-200)

A=13x1022 molecules

hope that helps

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