A container has in it 20.0 g of Ar and 10.0 of Ne. Thepressure in the container
ID: 682688 • Letter: A
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 VThe 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)Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.