Answer:
[tex]V_2=20.1L[/tex]
Explanation:
Hello there!
In this case, since this is a problem in which the pressure of the gas remains constant, we can use the Charles' law as a directly proportional relationship between temperature (in Kelvins) and volume given by:
[tex]\frac{V_2}{T_2} =\frac{V_1}{T_1}[/tex]
Thus, solving for the final volume as the temperature is decreased to -80.00 °C, we obtain:
[tex]V_2=\frac{V_1T_2}{T_1}[/tex]
[tex]V_2=\frac{30.0L*(-80+273)K}{(15+273)K} \\\\V_2=20.1L[/tex]
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