Question

At what temperature Q. at is the root mean square speed of ${N_2}$gas is equal to the propane gas at STP?
A. ${173.7^ \circ }C$
B. $173.7 K $
C. $ 273 K $
D. $ - {40^ \circ }C$

Answer 1

Hint:
Root mean square speed is defined as the square root of the mean of the squares of the speeds of all the molecules present in the given sample of the gas. We will understand the concept widely in the solution.

Complete step by step solution:
According to kinetic theory of gases we know that gas molecules are in a continuous random motion with high velocities and there is no loss of kinetic energy during collision as the collision is perfectly plastic. Therefore kinetic gas equation for one mole of a gas is given by
$PV = \dfrac{1}{3}m{N_A}{c^2}$ where c is the molecular speed

Root mean square speed is the hypothetical speed possessed by all the gas molecules when total kinetic energy is equally distributed amongst them. Total kinetic energy of a sample containing n number of molecules is
$ = \dfrac{1}{2}mc_1^2 + \dfrac{1}{2}mc_2^2 + .... + \dfrac{1}{2}mc_n^2$ (Where mass of every molecule is m)

If velocity possessed by all the molecules is same and that is c. then the total kinetic energy is equal to
= $n \times \dfrac{1}{2}m{c^2}$ . Equating both the expressions of total kinetic energies we will get
$n \times \dfrac{1}{2}m{c^2}$$ = \dfrac{1}{2}mc_1^2 + \dfrac{1}{2}mc_2^2 + .... + \dfrac{1}{2}mc_n^2$
now take $\dfrac{1}{2}m $ common from both sides and it will get cut on both the sides
$
  {c^2} = \dfrac{{c_1^2 + c_2^2 + .... + c_n^2}}{n} \\
  c = \sqrt {\dfrac{{c_1^2 + c_2^2 + .... + c_n^2}}{n}} \\
$

The value of c can be determined from the kinetic gas equation $PV = \dfrac{1}{3}m{N_A}{c^2}$. We get
$c = \sqrt {\dfrac{{3PV}}{M}} = \sqrt {\dfrac{{3RT}}{M}} $ (remember ideal gas equation PV=nRT)

Now according to the question let us suppose that root mean square speed of nitrogen is equal to that of propane gas we know that M(molar mass) of nitrogen is 28amu and of propane is 44amu.ryderberg constant will have the same value for both the gases since it is a universal constant. Substituting the values in the formula derived we get
${c_{{N_2}}} = {c_{{C_3}{H_8}}}$ $(\sqrt {\dfrac{{3R{T_{{N_2}}}}}{{28}}} = \sqrt {\dfrac{{3R{T_{{C_3}{H_8}}}}}{{44}}} )$

Squaring both sides, square root is removed, 3R gets canceled from both the sides and we get
$\dfrac{{{T_{{N_2}}}}}{{{T_{{C_3}{H_8}}}}} = \dfrac{{28}}{{44}} = \dfrac{7}{{11}} = 0.636$

Since the conditions given are STP so the temperature will be 300K and ${27^ \circ }C$ which is the temperature of propane gas. On substituting the values in the above equation we get
$
  {T_{{N_2}}} = 0.636 \times {T_{{C_3}{H_8}}} = 0.636 \times 273 = 173.7K \\
   = 0.636 \times 27 = {17.18^ \circ }C \\
$

Hence the correct option is option B.

Note:
There are three types of molecular speeds :Root mean square speed, Most probable speed and average speed.