Question

A litre of hydrogen at NTP weighs 0.09 g. Calculate R of this gas. What is the mass of a litre of this gas at 57 degree celsius and pressure of 86cm of Hg?

Answer 1

Hint: At NTP, the temperature of the gas is equal to 0 degree Celsius and the pressure is equal to 1 atm. Convert these in SI units. Number of moles of a gas is defined as the given mass of the gas divided by the molar mass of the gas. Then by using the ideal gas equation find the value of R. Using the same equation find the mass of the gas in the second question.
Formula used:
PV = nRT
$n=\dfrac{w}{M}$

Complete answer:
It is said that a gas of hydrogen at NTP weighs 0.09g and we have to calculate the value of R (universal gas constant). For this we will use the gas equation PV=nRT …… (i),
where P and V are the pressure and volume of the gas at temperature T, n are the number of moles of the gas.
At NTP (i.e. normal pressure and temperature), the pressure is equal to 1 atmosphere and the temperature is equal to ${{0}^{\circ }}C$.
$1atm=1.013\times {{10}^{5}}Pa$ and ${{0}^{\circ }}C=273K$.
Therefore, $P=1.013\times {{10}^{5}}Pa$, T = 273K and $V=1L={{10}^{-3}}{{m}^{3}}$.
The number of moles of a gas is equal to the given mass of the gas divided by the molar mass of the gas. i.e. $n=\dfrac{w}{M}$.
The molar mass of hydrogen gas is 2g.
$n=\dfrac{w}{M}=\dfrac{0.09}{2}=0.045$.
Substitute the values known values in equation (i).
$\Rightarrow 1.013\times {{10}^{5}}\times {{10}^{-3}}=0.045R\times 273$
$\Rightarrow R=\dfrac{1.013\times {{10}^{2}}}{0.045\times 273}=0.83\times {{10}^{2}}=8.3J{{K}^{-1}}mo{{l}^{-1}}$.
In the second question, it is asked to find the mass of one litre of the gas at a temperature of ${{57}^{\circ }}C$ and pressure of 86 mmHg.
We will use equation (i) and find the number of moles of the gas.
Now, $T={{57}^{\circ }}C=330K$, $P=86mmHg=1.14\times {{10}^{5}}Pa$ and $V=1L={{10}^{-3}}{{m}^{3}}$.
Substitute the values in equation (i).
$\Rightarrow 1.14\times {{10}^{5}}\times {{10}^{-3}}=n\times 8.3\times 330$
$\Rightarrow n=\dfrac{1.14\times {{10}^{2}}}{8.3\times 330}=0.041moles$.
We know that $n=\dfrac{w}{M}$
$\Rightarrow 0.041=\dfrac{w}{2}$
$\Rightarrow w=0.082g$.
Therefore, the mass of the hydrogen gas at the given temperature and pressure is 0.082g.

Note:
Note that there is a difference between molecular mass and molar mass.
Molecular mass is the mass of one molecule of the gas. Whereas molar mass is the mass of one mole of the same gas.
Molecular mass is measured in a.m.u and the molar mass is measured in grams.