Helium occupies 2.40 L at 123°C. What would the volume of the same sample of gas be at 23°C?

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Answer 1 · 2016-06-11T19:25:29

$V= 1.79 \ L$

$(P_1xxV_1)/T_1=(P_2xxV_2)/T_2$

Note that the pressure was not mentioned and therefore I would assume it remains constant.

$(cancel (P_1)xxV_1)/T_1=(cancel(P_2)xxV_2)/T_2$

$V_1/T_1=V_2/T_2$

$V_2=T_2/T_1xxV_1$

$V_2= (296 \ K)/(396 \ K) xx 2.40 \ L$

$V_2= (296 \ cancel (K))/(396 \ cancel (K)) xx 2.40 \ L$

$V_2= 1.79 \ L$