A $0.275L$ volume of an unspecified gas exerts a pressure of $732.6mm*Hg$ at a temperature of $-28.2$ $""^@C$ . What is the molar quantity of this gas?

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Answer 1 · 2017-04-03T23:31:33

Approx. $0.013*mol$

We need to know that $1*atm$ will support a column of mercury $760*mm$ high, and thus we can use the length of a mercury column to express pressure.

And thus $P=(732.6*mm*Hg)/(760*mm*Hg*atm^-1)=??*atm$

And we use the $"Ideal Gas Law"$ , $n=(PV)/(RT)$

$n=(PV)/(RT)=((732.6*mm*Hg)/(760*mm*Hg*atm^-1)xx0.275*L)/(0.0821*L*atm*K^-1*mol^-1xx245*K)=$

$1.32xx10^-2*mol.$

Why did I change the temperature to $"degrees Kelvin"$ ?

A 0.275*L0.275*L volume of an unspecified gas exerts a pressure of 732.6*mm*Hg732.6*mm*Hg at a temperature of -28.2-28.2 ""^@C""^@C. What is the molar quantity of this gas?