Question

The amount of heat required to increase the temperature of $1mol$ of a triatomic gas (non-linear) at constant volume is $n$ times the amount of heat required for $1mol$ of monatomic gas. The value of $n$ will be
\left( A \right)1 \\\ \left( B \right)1.3 \\\ \left( C \right)2 \\\ \left( D \right)2.5 \\\

Answer 1

Hint : In order to solve this question, we are going to calculate the values of the amount of heat required to increase the temperatures for the two types of gases by varying the values of the specific heats. After calculating the values, the ratio is calculated for the two to find the value of $n$ .

Complete Step By Step Answer:
When the volume is constant,
In order to increase the temperature of the $1mol$ of a gas, which is triatomic, the amount of heat required
${Q_1} = n{C_V}\Delta T$
Where $n$ the number of moles is ${C_V}$ is the specific heat at the constant volume, $\Delta T$ is the change in the temperature.
Now, putting the values of these quantities
${Q_1} = 1 \times 3R \times 1 = 3R$
Now for the monatomic gas, the amount of heat required to increase the temperature of $1mol$ of a gas,
${Q_2} = n{C_V}\Delta T$
Putting the values of the quantities, we get
${Q_2} = 1 \times 1.5R \times 1 = 1.5R$
As it is given that the amount of heat required to increase the temperature of $1mol$ of a gas. Now calculating the ratio of the heats for the two types of gases.
Thus, $\dfrac{{{Q_1}}}{{{Q_2}}} = 2$
Hence, option $\left( C \right)2$ is the correct answer.

Note :
The amount of heat required to increase the temperature of one mole of a gas for a triatomic gas is twice the amount of heat required to increase the temperature of the one mole for the monatomic gas. These two values depend upon the specific heat of the two types of gases.