Question

What amount of heat must be supplied to $2.0\times {{10}^{-2}}kg$ of nitrogen (At room temperature) to raise its temperature by ${{45}^{0}}C$ at constant pressure?
(Molecular mass of ${{N}_{2}}=28$, $R=8.3J.mo{{l}^{-1}}{{K}^{-1}}$)

Answer 1

Hint : This problem can be solved by finding out the number of moles of the gas given, its specific heat capacity at constant pressure and using the formula for the heat required to change the temperature of a gas in terms of its specific heat capacity at constant pressure, number of moles and the temperature change.
Formula used:
$n=\dfrac{m}{M}$
${{C}_{P}}=\dfrac{7R}{2}$
$Q=n{{C}_{p}}\Delta T$

Complete step by step solution :
We will find out the number of moles, its molar specific heat capacity at constant pressure and apply the direct formula for the heat required for the temperature change, using these two values.
Therefore, let us analyze the question.
The mass of the given nitrogen gas is $m=2\times {{10}^{-2}}kg$.
The molecular mass of ${{N}_{2}}$ is $M=28g/mol$.
Let the number of moles of the nitrogen gas be $n$.
Since, the process is being done at constant pressure, we will find out the specific molar heat capacity at constant pressure of nitrogen gas.
Let the specific molar heat capacity at constant pressure of nitrogen gas be ${{C}_{P}}$.
Let the heat required for the process be $Q$.
The temperature change as given in the question is $\Delta T={{45}^{0}}C=45K$
$\left( \begin{aligned} & {{\because }^{0}}C=273+K \\\ & \therefore {{\Delta }^{0}}C=\Delta \left( 273+K \right)=\Delta 273+\Delta K=0+\Delta K=\Delta K \\\ \end{aligned} \right)$
That is, a change in temperature in Celsius is the same as a change in temperature in Kelvin.
It is also given that the universal gas constant $R=8.3J.mo{{l}^{-1}}{{K}^{-1}}$.
Now, the number of moles $n$ of a mass $m$ of gas with molecular mass $M$ is given by
$n=\dfrac{m}{M}$ --(1)
Therefore, using (1), we get
$n=\dfrac{m}{M}$
$\therefore n=\dfrac{2\times {{10}^{-2}}kg}{28g/mol}=\dfrac{2\times {{10}^{-2}}kg}{28\times {{10}^{-3}}kg/mol}=0.71mol$ $\left( \because 1g={{10}^{-3}}kg \right)$
Also, the specific molar heat capacity at constant pressure ${{C}_{P}}$ of a diatomic gas is given by
${{C}_{P}}=\dfrac{7R}{2}$ ---(2)
where $R=8.3J.mo{{l}^{-1}}{{K}^{-1}}$ is the universal gas constant.
Since nitrogen is also a diatomic gas $\left( {{N}_{2}} \right)$, using (2), we get,
${{C}_{P}}=\dfrac{7R}{2}$
Now, the heat required $Q$ to increase the temperature by $\Delta T$ of an ideal gas with $n$ moles at constant pressure is given by
$Q=n{{C}_{P}}\Delta T$ --(3)
Where ${{C}_{P}}$ is the molar specific heat capacity at constant pressure of the gas.
Using (3), we get
$Q=n{{C}_{P}}\Delta T$
$\therefore Q=0.71\times \dfrac{7R}{2}\times 45$
$\therefore Q=0.71\times \dfrac{7}{2}\times 8.3\times 45=928.15J$
Hence, the amount of heat required for the process is $928.15J$.

Note : Students must be careful and check in which units the temperature is given in thermodynamics problems. The temperatures in all thermodynamics problems must be converted into the Kelvin unit since it is the standard SI unit of temperature and all thermodynamic equations are valid only for this unit. Sometimes questions are purposefully set in such a way so that students do not notice that the temperatures given are in some other units and they commit a silly mistake.