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. The volume is V. 5.0 g of Ne is removed from this container. Temperatureremains the same. To raise the pressure back to 1000 torr,the volume must be changed from V to a. 1.20 V b. 0.80 V c. 0.67 V d. 0.75 V
The answer is D but why? A container has in it 20.0 g of Ar and 10.0 of Ne. Thepressure in the container is 1000 torr. The volume is V. 5.0 g of Ne is removed from this container. Temperatureremains the same. To raise the pressure back to 1000 torr,the volume must be changed from V to a. 1.20 V b. 0.80 V c. 0.67 V d. 0.75 V
The answer is D but why?

Explanation / Answer

         Thenumber of moles of Ar in the mixture = 20.0 g /39.95g/mol                                                                          = 0.5006         The number ofthe moles of Ne = 10.0 g / 20.18g/mol                                                          = 0.4955 mol        total number ofmoles = 0.9961 mol        Pressure = 1000torr        Volume V1 = V       The number of molesof Ne present in 5.0 g   = 5.0 g / 20.18g/mol                                                                              = 0.2477 mol       0.2477 moles isremoved from the mixture .      Number of moles present in themixture = 0.9961 - 0.2477                                                                n2   = 0.7484 mol      The temperature remains constantand the pressure is also same in the initial and final states.             PV = nRT          According toAvagadro's law           (V1 / n1 ) = V2 / n2                   (V / 0.9961) = V2 /0.7484                    V2 = 0.75 V        The answer is d)                   (V / 0.9961) = V2 /0.7484                    V2 = 0.75 V        The answer is d)        

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