Question

The density of hydrogen gas at STP is $0.09kg{m^{ - 3}}$. The mean kinetic energy of one gram molecule of hydrogen gas is approx.-
(A) $3403\;J$
(B) $3500\;J$
(C) $3704\;J$
(D) $3207\;J$

Answer 1

Hint: To calculate the mean kinetic energy of the hydrogen gas, we first calculate the root mean square velocity of the gas. We can calculate it by dividing the density of hydrogen gas by three times the given pressure, which is equal to $1\;atm$ here.
Formula used:
${v_{rms}} = \sqrt {\dfrac{{3P}}{\rho }} $

Complete step by step solution:
In the question, it is given to us that,
The density of hydrogen gas is, $\rho = 0.09kg{m^{ - 3}}$
The pressure at STP is equal to, $P = 1.01 \times {10^5}Pa$
(All values should be in SI units to get the answer in SI units)
For a gas, the root mean squared or RMS speed is given by the formula-
${v_{rms}} = \sqrt {\dfrac{{3P}}{\rho }} $
For the given hydrogen gas, the RMS speed is,
${v_{rms}} = \sqrt {\dfrac{{3 \times 1.01325 \times {{10}^5}}}{{0.09}}} $
$ \Rightarrow {v_{rms}} = \sqrt {\dfrac{{3 \times 1.01325 \times {{10}^5}}}{{0.09}}} $
On solving further, we get-
${v_{rms}} = \sqrt {\dfrac{{3.03975 \times {{10}^5}}}{{0.09}}} $
$ \Rightarrow {v_{rms}} = \dfrac{{551.33}}{{0.3}} = 1837.79$
The root mean square velocity of the given hydrogen gas is $1837.79m/s$.
The next step is to calculate the mass of $1$ gram molecule of hydrogen.
A gram molecule of a gas is the mass in grams which is equal to the corresponding molar mass of that substance, therefore for hydrogen gas, it will represent the mass of one mole of the gas.
We know that at STP,
The volume occupied by $1$ mole of ${H_2}$gas is $V = 22.4L$
We know that, $1L = {10^{ - 3}}{m^3}$
Thus, $V = 22.4 \times {10^{ - 3}}{m^3}$
It is given in the question that the density of the hydrogen gas is $\rho = 0.09kg/{m^3}$
Then the mass of the hydrogen gas is given by,
$M = \rho V$
$ \Rightarrow M = 0.09 \times 22.4 \times {10^{ - 3}}kg$
$ \Rightarrow M = 2.016 \times {10^{ - 3}}kg$
The kinetic energy for $1$ gram molecule is given by-
$K = \dfrac{1}{2}Mv_{rms}^2$
On putting the values we obtain-
$K = \dfrac{1}{2} \times 2.016 \times {10^{ - 3}} \times {\left( {1837.79} \right)^2}$
$K = 3399.425J$

The kinetic energy of the molecules is $3399.43\;J$, the closest answer is option (a), therefore it is the correct answer.

Note: For questions that have large calculations and involve operations such as squaring- square root, fractions-multiplication, and logarithms, the calculations must be as precise as possible because even a small approximation may change the answer significantly, to get the most accurate answer, more number of digits must be taken after the decimal.