Question

One Torr is equal to
(A) 1 cm of Hg
(B) 1 atm pressure
(C) 1 mm of Hg
(D) 1 $Nm^{-2}$

Answer 1

Hint:Torr is a unit of pressure named after the scientist Evangelista Torricelli, who discovered a barometer's principle. Torr was named in honor of the inventor of the barometer, Evangelista Torricelli. Torr is used to represent minimal pressure values.

Complete step by step answer:
Torr is a unit of pressure. It is defined as precisely $\dfrac{1}{{760}}$of a standard atmosphere.
It can be written in the formula as,
$1\;{\rm{torr = }}\dfrac{1}{{760}} \times {P_{atm}}\;......\;\left( {\rm{I}} \right)$
Here ${P_{atm}}$ is the standard atmospheric pressure.
Since we know that $1\,{P_{atm}} = 760\;{\rm{mm}}\;{\rm{Hg}}$ and we will substitute this in equation (I) and we get,
$\begin{array}{l}
1\;{\rm{torr = }}\dfrac{{760\;{\rm{mm}}\;{\rm{Hg}}}}{{760}}\\
 = 1\;{\rm{mm}}\;{\rm{Hg}}
\end{array}$
Therefore option (C) is correct.
We will now convert 1 mm Hg into cm of Hg.
$\begin{array}{l}
1\;{\rm{torr = }} = 1\;{\rm{mm}}\;{\rm{Hg}} \times \dfrac{{{{10}^{ - 1}}\;{\rm{cm}}\;{\rm{Hg}}}}{{1\;{\rm{mm}}\;{\rm{Hg}}}}\\
 = 0.1\;{\rm{cm}}\;{\rm{Hg}}
\end{array}$
Therefore, option (A) is wrong.
We know that 1 torr is defined as $\dfrac{1}{{760}}$of a standard atmosphere as mentioned earlier that is,
$1\;{\rm{torr = }}\dfrac{1}{{760}} \times {P_{atm}}$
Therefore, option (B) is wrong.
We know that $1\,{P_{atm}} = 101325\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}$, and we will substitute this value in equation (I) and we will get,
$\begin{array}{l}
1\;{\rm{torr = }}\dfrac{{101325\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}}}{{760}}\\
 = 133.322\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}
\end{array}$
Therefore, option (D) is also wrong.
Therefore, the correct option is $1\;{\rm{torr}} = 1\;{\rm{mm}}\;{\rm{Hg}}$, and the right option is (C).

Additional information:
 Pressure (p) is the force applied perpendicular to an object's surface per unit area. Some other units of pressure are below. The bar (denoted by the bar), defined as 100 kPa precisely. The atmosphere (indicated by atm), defined as 101.325 kPa.

Note: Earlier, one Torr was intended to be the same as one "millimeter of mercury," but subsequent research of the two units made them slightly different. It is one of the measuring units in a high vacuum. Some devices, such as Electron beam machining, operate at minimal pressures near to vacuum, and their Torr as a pressure unit is used for measurement.