Question

The H calories of heat is required to increase the temperature of one mole of monoatomic gas from${{20}^{0}}$C to${{30}^{0}}$C at constant volume. The quantity of heat required to increase the temperature of 2 moles of a diatomic gas from${{20}^{0}}$C to${{25}^{0}}$C is at constant volume is:
$A.\quad \dfrac{4H}{3}$
$B.\quad \dfrac{5H}{3}$
$C.\quad 2H$
$D.\quad \dfrac{7H}{3}$

Answer 1

Hint: The question requires knowledge of thermodynamics, more specifically the conditions governing the mechanism of isochoric process (process of constant volume), which is, $dQ=dU\Rightarrow dQ=n{{C}_{V}}dT.$ The different values of specific heat for monoatomic gas, diatomic gas, triatomic gas and so on, is also required to solve this problem.

Complete step by step solution:
Let’s start by deriving the relations that we will get in between various quantities during the isochoric process. As we know, this process occurs when the volume of the body remains constant throughout the process. $\Rightarrow \Delta V=0.$ We know that, $dQ=dU+dW$and$dW=PdV.$
Therefore, $dQ=dU+PdV$ and since $dV=0,$ therefore, $dQ=dU.$
The value of internal energy$(U)$is, $U=n{{C}_{V}}dT$.
Hence, $dU=n{{C}_{V}}dT$. Substituting in this value into the above equation for energy, $dQ=dU\Rightarrow dQ=n{{C}_{V}}dT.$
Let’s consider the first case now:
Here, we have one mole of monoatomic gas, having a change in temperature change of ${{10}^{0}}$C. All of this happening under isochoric condition. The amount of heat required is H calories.
Therefore,${{n}_{1}}=1,\Delta T={{10}^{0}}C=10K$. Further, for monoatomic gas, the specific heat is given by,${{C}_{V}}=\dfrac{3}{2}R$
Using these in $d{{Q}_{1}}={{n}_{1}}{{C}_{V}}dT\Rightarrow H=1(\dfrac{3R}{2})10\Rightarrow H=15R.$
That is, $R=\dfrac{H}{15}.$
Now, let’s consider the next case:
Here, we have 2 moles of a diatomic gas, where the change in temperature is ${{5}^{0}}$C. Again, the condition remains the same, that is, isochoric condition.
Therefore, ${{n}_{2}}=2,\Delta T={{5}^{0}}C=5K.$ Further, for monoatomic gas, the specific heat is given by, ${{C}_{V}}=\dfrac{5}{2}R$
Hence, $d{{Q}_{2}}={{n}_{2}}{{C}_{V}}dT\Rightarrow d{{Q}_{2}}=2(\dfrac{5R}{2})5\Rightarrow d{{Q}_{2}}=25R.$
Substituting in the value of R from the previous case into the above equation, we get, $d{{Q}_{2}}=25(\dfrac{H}{15})\Rightarrow d{{Q}_{2}}=\dfrac{5H}{3}.$
Hence, the amount of heat required in this case is $\dfrac{5H}{3}.$

Note: The value of internal energy contains specific heat in constant volume denoted as ${{C}_{V}}$. This is because the process that we are considering has the volume value constant throughout the process. Similarly, for the process of pressure constant process, the specific heat is given by ${{C}_{P}}$.