# What will be unit of volume in V=nRT/p if p is in Torr instead of atm? [closed]

$$V=nRT/p$$, where $$n$$ is amount of gas in $$\mathrm{mol}$$, $$R$$ is gas constant, $$V$$ is volume in $$\pu{dm3}$$, and $$p$$ is pressure in $$\pu{atm}$$.

What happens to $$R$$ if $$p$$ is in $$\pu{Torr}$$ instead of $$\pu{atm}$$?

Does this question also have some relation with my question above in title (What will be unit of volume in $$V=nRT/p$$ if $$p$$ is in $$\pu{Torr}$$ instead of $$\pu{atm}$$?)? If yes, please answer this too.

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• You make up your own unit or you use the gas constant in the appropriate units. Either way, it is way easier to convert from torr to Pa or bar. – TAR86 Mar 10 at 8:07

Regarding the universal gas constant $$R$$, you may find it expressed in multiple units -- and consequentially, sometimes numerically quite different -- in textbooks about Physical Chemistry, or even right on top right of the corresponding entry of wikipedia.

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Regarding your second point, of course you may use cubic metres as unit of volume, or non-SI but still SI accepted units like the litre (cf. here). What you shall do then (a check useful even if using SI units only, too) is to perform a dimensional analysis. Simply speaking, your equation is true if not only the numbers, but equally the units on the left, and on the right side of the equation sign match each other. For the units, though, some of them may cancel out each other if related by multiplication and division, as a later example of the second reference (accidentally about the Universal Gas Law), too:

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What happens to $$\mathrm{R}$$ if $$\mathrm{P}$$ is in $$\pu{Torr}$$ instead of $$\pu{atm}$$?

Short answer: you can use the value of $$\mathrm{R}= \pu{62.32~Torr\cdot{L}\cdot{K^{-1}}\cdot{mol^{-1}}}$$ and the unit of $$\mathrm{V}$$ in litere. These are gas constants $$\mathrm{R}$$ and each is used with the appropriate units. \begin{align} \mathrm{R}&=\frac{\mathrm{PV}}{\mathrm{T\cdot{n}}}\\ &=\frac{\pu{atm}\times{22.4L}}{\pu{273K\cdot{mol}}}\\ &=\pu{0.082atm\cdot{L}\cdot{K^{-1}}\cdot{mol^{-1}}}\\ &=\pu{0.082\times{\pu{760Torr}}\cdot{L}\cdot{K^{-1}}\cdot{m^{-1}}}\\ &=\pu{62.32~Torr\cdot{L}\cdot{K^{-1}}\cdot{mol^{-1}}}\\ &=\pu{62.32~mlHg\cdot{L}\cdot{K^{-1}}\cdot{mol^{-1}}}\\ &=(\pu{62.32\times{10^{-3}~m})(\pu{1.35951\times{10^4Kg/m^3}})(\pu{9.81m/s^2}})\pu{K^{-1}\cdot{mol^{-1}}}\\ &=\pu{8.31\times{10^3}\times{\frac{Kg\cdot{m}}{m^2\cdot{s^2}}}}\times{\pu{L\cdot{K^{-1}}\cdot{mol^{-1}}}}\\ &=\pu{8.31\times{10^3\times{\frac{N}{m^2}}}}\times{\pu{L\cdot{K^{-1}}mol^{-1}}}\\ &=8.31\times{10^3}~\pu{Pa}\cdot{L}\cdot{K^{-1}}\pu{mol^{-1}}\\ &=\pu{8.31KPa}~.\pu{L\cdot{K^{-1}}mol^{-1}}\\ &=\pu{8.31J}~.\pu{K^{-1}mol^{-1}}\\ &=\pu{1.98Cal.}~\pu{K^{-1}mol^{-1}}\\ \end{align}