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Sagot :
Answer:
a. V=44.9mL
Explanation:
For a gas undergoing all of these changes, it will be important to combine Boyle's Law and Charles' Law to form the following equation (if it isn't already known):
[tex]\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}[/tex]
Where the Temperatures must be measured in Kelvin.
Recall that to convert Celsius to Kelvin, one must add 273 or use the equation [tex]T_C+273=T_K[/tex].
Thus, [tex]T_1=(18.1+273)[K]=291.1[K][/tex] and [tex]T_2=(12.5+273)[K]=285.5[K][/tex]
To solve for the requested quantity, note that all of the other units match between beginning and end, so we substitute and solve:
[tex]\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}[/tex]
[tex]\dfrac{(2.61[atm])V_1}{(291.1[K])}=\dfrac{(1.92[atm])(59.9[mL])}{(285.5[K])}[/tex]
[tex]\dfrac{(2.61[atm] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----})\bold{V_1}}{291.1[K] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----}}*\dfrac{291.1[K] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----}}{2.61[atm]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----}}=\dfrac{(1.92[atm]\!\!\!\!\!\!\!\!\!\!\!{--})(59.9[mL])}{285.5[K\!\!\!\!\!{-}]}*\dfrac{291.1[K\!\!\!\!\!{-}]}{2.61[atm]\!\!\!\!\!\!\!\!\!\!\!{--}}[/tex]
[tex]V_1=44.928677576[mL][/tex]
Accounting for significant digits, [tex]V_1=44.9[mL][/tex]
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