# 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.

• 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 '19 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. 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}