Question

When we heat a gas sample from ${{27}^{0}}C$ to ${{327}^{0}}C$, the initial average kinetic energy of the gas molecules was $E$. What will be the average kinetic energy after heating?
$A)\text{ }\sqrt{2}E$
$B)\text{ 2}E$
$C)\text{ 300}E$
$D)\text{ 327}E$

Answer 1

This problem can be solved by using the fact that the average kinetic energy of the gas molecules of an ideal gas is directly proportional to its absolute temperature. By using this proportionality theorem, we can get the final average kinetic energy as a function of the initial kinetic energy using the initial and final absolute temperatures.
Formula used:
$E\propto T$

Complete answer:
We will use the fact that the average kinetic energy $E$ of the gas molecules of an ideal gas is directly proportional to the absolute temperature $T$ of the gas.
$\therefore E\propto T$
$\therefore E=\lambda T$ --(1)
where $\lambda$ is the constant of proportionality.
Now, let us analyze the question.
The initial absolute temperature of the gas is ${{27}^{0}}C=\left( 27+273 \right)K=300K$ $\left( \because {{T}^{0}}C=\left( 273+T \right)K \right)$
The final temperature of the gas is ${{327}^{0}}C=327+273=600K$ $\left( \because {{T}^{0}}C=\left( 273+T \right)K \right)$
The initial average kinetic energy of the gas molecules is $E$.
Let the final average kinetic energy of the gas molecules be ${{E}_{f}}$.
Now, using (1), we get
$E=\lambda 300$ --(2)
${{E}_{f}}=\lambda 600$ --(3)
Therefore, using (2) and (3), we get
$\dfrac{{{E}_{f}}}{E}=\dfrac{\lambda 600}{\lambda 300}=\dfrac{600}{300}=2$
$\therefore {{E}_{f}}=2E$
Hence, we have got the average kinetic energy of the gas molecules after heating as $2E$.

Therefore, the correct option is $B)\text{ 2}E$.

Note:
Students must note that while solving problems of thermodynamics they must always convert each and every temperature value given to the Kelvin scale first and then use these values for their calculations and in thermodynamic formulas and equations. This is because the thermodynamics equations are valid only for the Kelvin scale temperature values. If we had proceeded using the Celsius scale values of the temperature as given in the question, we would have arrived at a completely different and wrong answer.