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The pressure of a sample of helium is 1.556 atm in a 268.5 mL container. If the container is compressed to 112.4 mL without changing the temperature, what is the new pressure?
a. 3.72
b. 0.651
c. 1.94e4
d. 277


Sagot :

Answer:

a.  3.72 [atm]

Explanation:

For a gas at constant temperature, (with no change in number of molecules of the gas), we can apply Boyle's Law:  [tex]P_1V_1=P_2V_2[/tex]

[tex](1.556[atm])(268.5[mL])=P_2(112.4[mL])[/tex]

[tex]\dfrac{(1.556[atm])(268.5[mL\!\!\!\!\!\!\!\!{--}])}{112.4[mL \!\!\!\!\!\!\!\!{--}]}=\dfrac{P_2(112.4[mL]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----})}{112.4[mL]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----}}[/tex]

[tex]3.716957[atm]=P_2[/tex]

It seems like the answer should have 4 significant figures since all of the other quantities have 4 significant figures, but the closest answer choice of those provided is a.  3.72