Question

$14g$ of ${{N}_{2}}$ gas is heated in a closed rigid container to increase its temperature from $23{}^\circ C$ to $43{}^\circ C$. The amount of heat supplied to the gas will be given as,
$\begin{align}
  & A.25cal \\
 & B.50cal \\
 & C.100cal \\
 & D.30cal \\
\end{align}$

Answer 1

Hint: First of all, the number of moles of gas inside the closed rigid container should be found out. In a closed rigid container, the change in volume of the gas will be zero. Therefore the work done will be also zero. Using these details the heat supplied is calculated which will be the product of number of moles, specific heat capacity at constant volume and the change in temperature. These all may help you to solve this question.

Complete step by step answer:
First of all let us calculate the number of moles in the container. The mass of the gas inside the container and the molecular mass of the nitrogen gas is also known. Therefore we can write that,
$\begin{align}
  & m=14g \\
 & M=28g \\
\end{align}$
Using these values we can calculate the number of moles which can be written as,
$n=\dfrac{m}{M}$
Substituting the values in it will give,
$n=\dfrac{14}{28}=\dfrac{1}{2}$
As the container is closed and rigid, the volume change will be zero. Therefore we can write that,
$\Delta V=O$
Hence the work done will be also zero as the volume change is zero. This can be given by the equation,
$\Delta W=P\Delta V=O$
Therefore according to the equation,
$\Delta Q=\Delta U+\Delta W$
As the change in work done is zero, the heat supplied will be equal to the change in internal energy. This can be shown as,
$\Delta Q=\Delta U$
The change in internal energy is given by the product of number of moles, specific heat capacity at constant volume and the change in temperature.
Therefore the equation will be,
$\Delta Q=n{{C}_{V}}\Delta T$
The nitrogen molecule is given as diatomic. Therefore the specific heat at constant volume can be expressed as,
${{C}_{V}}=\dfrac{5R}{2}$
Substituting the values in the equation will give,
$\Delta Q=n\dfrac{5R}{2}\Delta T$
Where $R$be the universal gas constant.
As the temperature difference is,
$\Delta T=43-23=20{}^\circ C$
And universal gas constant is,
$R=2\dfrac{cal}{mol-{}^\circ C}$
We can write that,
$\Delta Q=\dfrac{1}{2}\times \dfrac{5}{2}\times 2\times 20=50cal$

So, the correct answer is “Option B”.

Note: Specific heat capacity is the amount of heat required to raise the temperature of one kilogram of substance by one kelvin. There are two specific heat capacities. One is specific heat capacity at constant volume, ${{C}_{V}}$. Another one is specific heat capacity at constant pressure, ${{C}_{P}}$.