Question

If the universal gas contain is 8.3$$$$joule$$$$mo{l^{ - 1}}{K^{ - 1}} and the Avogadro’s number is $6 \times10 {\^{23}}$. The mean kinetic energy of oxygen molecules at ${327^ \circ }C$ will be:
A. $415 \times {10^{ - 23}}joule$
B. $2490 \times {10^{ - 24}}joule$
C. $1245 \times {10^{ - 23}}joule$
D. $830 \times {10^{ - 22}}joule$

Answer 1

The kinetic energy of a gas is a measurement of its temperature, changing of temperature causes the molecule to increase their movement this rise to kinetic energy. We apply this formula to solve this problem, K.E. $= \dfrac{{3RT}}{{2{N_A}}}$ where, $(R)$ is universal gas constant, ${N_A}$is the numbers of mole, $T$ is known as temperature of gases and also convert the temperature of from degree Celsius $\left( {^ \circ } \right)$ to kelvin $\left( K \right)$ by using $\left( {T + 273} \right)K$ formula.

Complete step-by-step solution:
Given, the universal gas constant (R)$$$$ = 8.3J/mol/K
Avogadro’s number $\left( {{N_A}} \right) = {6.10^{23}}$
As we know there are so many methods of temperature conversion but here we use the Kelvin method which is $Degree\,Celsius = \left( {T + 273K} \right)$. In the Kelvin scale the boiling point is $373.15K$ and the freezing point is $273.15K$.
In this question temperature, $\left( T \right) = {327^ \circ }$
This can be also written as-

$$T = 327 + 273K \\\ = 600K \\\$$

Therefore, the mean kinetic energy of oxygen molecules at ${327^ \circ }C$ will be

$$KE = \dfrac{{3RT}}{{2{N_A}}} \\\ = \dfrac{{3 \times 8.3J/mol/K \times (600K)}}{{2 \times 6 \times {{10}^{23}}}} \\\ = 1245 \times {10^{ - 23}}joule \\\$$

So, option c) $1245 \times {10^{ - 23}}joule$ is correct.

Note: The increase of kinetic energy directly depends on the temperature. Total sum of kinetic energy is known as heat.