Two glass bulbs are connected by a valve. one bulb has a volume of 650.0 milliliters and is occupied by n2(g) at 825 torr. the other has a volume of 500.0 milliliters and is occupied by o2(g) at 732 torr. the valve is opened and the two gases mix. calculate the total pressure and the partial pressures of n2(g) and o2(g) in the resulting mixture.

First we are going to use boyle's law:

1- gas (N2) only : initially it is contained in Volume = 0.65 L .and finally it will spread out into V = 1.15 L when the initial pressure was 825 torr

the final pressure will be V1 * P1 = V2 * P2

0.65L x 825 torr = 1.15L x P2

So, P2 = 0.65L x 825 torr / 1.15L = 466.3 torr

∴ the partial pressure of N2 gas = 466.3 torr

2- then now O2 gas only: initially it is contained in V = 0.5L .and finally it will spread out into V = 1.15L when its initial pressure was 732 torr

so, the final pressure will be V1 * P1 = V2 * P2

0.5L x 732 torr = 1.15L x P2

∴ P2 = 0.5L x 732 torr / 1.15L = 318.3 torr

∴ the partial pressure of O2 gas = 318.3 torr.

3-The total pressure of both gases spread in the two bulbs is:

= the partial pressure of N2 + the partial pressure of O2

= 466.3 + 318.3 = 784.6 torr

∴ The total pressure is = 784.6 torr

m=9

step-by-step explanation:

b. examine and describe cells