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

Hint: 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.