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If a gas occupies 1532.7 mL at standard temperature, what volume does it occupy at 49.4 ºC if the pressure remains constant?
a. 1810
b. 1300
c. 276
d. 3.80e5


Sagot :

Answer:

a. 1810mL

Explanation:

When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:

[tex]\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}[/tex]  where the temperatures must be measured in Kelvin

To convert from Celsius to Kelvin, add 273, or use the equation:  [tex]T_C+273=T_K[/tex]

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, [tex]T_1 = 273[K][/tex], and [tex]T_2 = (49.4+273)[K]=322.4[K][/tex].

[tex]\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}[/tex]

[tex]\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}[/tex]

[tex]\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})[/tex]

[tex]1810.04571428[mL]=V_2[/tex]

Adjusting for significant figures, this gives [tex]V_2=1810[mL][/tex]

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