Question

How would you solve the equation for U for ${\text{R}}$?

Answer 1

Hint: For solving the above given equation for universal gas constant (${\text{R}}$) we have to do rearrangement on the above given equation and also get the measuring unit of ${\text{R}}$ by canceling opposite units of others.

Complete answer:
In the question an equation is given which is known as an ideal gas equation and represent as-
${\text{PV = nRT}}$
Where, ${\text{P}}$ = Pressure of gas (in ${\text{atm}}$),
${\text{V}}$= Volume of gas (in ${\text{L}}$),
${\text{n}}$ = No. of moles of gas (in ${\text{mol}}$),
${\text{T}}$ = Temperature of gas (in ${\text{K}}$) and
${\text{R}}$ = Universal gas constant = to find?
Now for calculating the value of universal gas constant (${\text{R}}$) we will rearrange the above given equation and get,
\[{\text{R = }}\dfrac{{{\text{P}}\left( {{\text{atm}}} \right){\text{V}}\left( {\text{L}} \right)}}{{{\text{n}}\left( {{\text{mol}}} \right){\text{T}}\left( {\text{K}} \right)}}\] ……. (i)
The measuring unit of universal gas constant is ${\text{atmLmo}}{{\text{l}}^{{\text{ - 1}}}}{{\text{K}}^{{\text{ - 1}}}}$.
Now we calculate the value of universal gas constant at standard temperature and pressure. So, we have to calculate the value of universal gas constant (${\text{R}}$) when temperature is ${\text{273K}}$, pressure is ${\text{1atm}}$for one mole of gas which takes up the volume of ${\text{22}}{\text{.4L}}$, according to the Avogadro’s Law.
By putting all these values in the equation (i) and we get,
\[{\text{R = }}\dfrac{{{\text{1}}\left( {{\text{atm}}} \right){\text{22}}{\text{.4}}\left( {\text{L}} \right)}}{{{\text{1}}\left( {{\text{mol}}} \right){\text{273}}\left( {\text{K}} \right)}}{\text{ = 0}}{\text{.0821atmLmo}}{{\text{l}}^{{\text{ - 1}}}}{{\text{K}}^{{\text{ - 1}}}}\]

Note:
Here some of you may do wrong calculations if you put different values in the value of temperature, pressure, moles and volume of gas with respect to the measuring units. Because if we do so then the obtained value of ${\text{R}}$ will be wrong.