A container has in it 20.0 g of Ar and 10.0 of Ne. Thepressure in the container

Question

A container has in it 20.0 g of Ar and 10.0 of Ne. Thepressure in the container is 1000 torr. After 5.0 g of Ne isremoved from this container, the pressure becomes (volume andtemperature remain the same) a. 650 torr b. 700 torr c. 750 torr d. 800 torr
I know the answer is C but how? A container has in it 20.0 g of Ar and 10.0 of Ne. Thepressure in the container is 1000 torr. After 5.0 g of Ne isremoved from this container, the pressure becomes (volume andtemperature remain the same) a. 650 torr b. 700 torr c. 750 torr d. 800 torr
I know the answer is C but how?

Explanation / Answer

From the ideal gas equation we have                   PV = nRT Where V, R, T are constants, thus we get P n 'Therefore the pressure in the container is proportionalto the number of moles of gas present in the compound. The number of moles of 'Ar' present = 20g/39.94g/mole                                                         =0.5 mole Similarly the number of moles of 'Ne' present =10.0g/20.17g/mole                                                                      = 0.495 mole Thus the total moles present initially = 0.5 mole+0.495 mole                                                        = 0.995 mole When we remove 5 g of Ne then The number of moles of 'Ne' present = 5.0g/20.17g/mole                                                         = 0.2478 mole Therefore the number of moles remained = 0.995mole - 0.2478 mole = 0.747 mole Therefore Let us assume P1 = 1000 torr                                       P2 = ?                                        n1 = 0.995 mole , n2 = 0.747 mole         P1/P2   = n1 /n2          1000 torr /P2 = 0.995 /0.747               P2 = 750 torr          1000 torr /P2 = 0.995 /0.747               P2 = 750 torr

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