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The valve between a 3.25-L tank containing O2(g) at 7.36 atm and a 3.30 L tank containing Ne(g) at 6

The valve between a 3.25-L tank containing O2(g) at 7.36 atm and a 3.30 L tank containing Ne(g) at 6.00 atm is opened. Calculate the ratio of partial pressures (O2:Ne) in the container. a. 0.93

b. 1.14

c. 1.21

d. 0.878

e. 0.551


Explanation:

Initial conditions: O2

V=3.25-L

P = 7.36 atm

Initial conditions: Ne

V= 3.30 L

P=6.00 atm

Final conditions:

Since the valve is opened, the new volume would be 3.25L + 3.30L = 6.55L


We can use P1V1=P2V2 for each gas to calculate the new pressure of each gas.

O2: 7.36 atm(3.25L) = P2(6.55L)

P2= 3.65 atm

Ne: 6.00 atm(3.30 L)=P2(6.55L)

P2 = 3.02 atm

Let's now do the ratio of O2 to Ne pressure. 3.65 atm/3.02 atm = 1.21


Answer: c

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