A $42.3mL$ volume of gas at $371.3K$ is cooled to $351.7*K$ . What is the new volume?

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Answer 1 · 2017-07-08T17:29:06

We get approx. $V_2<=50*mL$

We use old $"Charles' Law"$ to give $VpropT$ for a given quantity of gas...........

Thus $V=kT$ , and solving for $k$ in each scenario.....

$V_1/T_1=V_2/T_2$ , ad of course $T$ is quoted in $"degrees Kelvin"$ .

And so we solve for $V_2=V_1/T_1xxT_2=(42.3*mLxx351.7*K)/(371.3*K)$

$=??*mL........$

A 42.3*mL42.3*mL volume of gas at 371.3*K371.3*K is cooled to 351.7*K351.7*K. What is the new volume?