Question

State first law of thermodynamics and derive the relation between molar specific heats of a gas.

Answer 1

For finding relation between ${C_P}\;{\text{and }}{C_V}\;$
Use the equation $q{\text{ }} = {\text{ }}n{\text{ }}C{\text{ }}\Delta T$, and apply for ${C_P}\;{\text{and }}{C_V}\;$.
And also apply the fact that:
At constant pressure, change in heat = change in enthalpy i.e. $Q = \Delta H$
At constant volume, change in heat = change in internal energy i.e. $Q = \Delta U$

Complete step by step answer:
The first law of thermodynamics is nothing but the application of the principle of conservation of energy to heat energy and thermodynamic processes and systems.The first law of thermodynamics states that the heat added to the system minus the work done by the system is equal to the change in internal energy of the system.The standard unit for all these quantities is joule (J), but they are sometimes also expressed in calories, where calorie = 4.2 joules.
Relation between molar specific heats:
There are two molar specific heat values for a gas:
The molar heat capacity C, at constant volume, is represented by ${C_V}$.
At constant pressure, the molar heat capacity C, is represented by ${C_P}$.
In the following section, we will find how ${C_P}\;{\text{and }}{C_V}\;$are related, for an ideal gas.
From the equation $q{\text{ }} = {\text{ }}n{\text{ }}C{\text{ }}\Delta T$, we can say:
At constant pressure P, we have
${q_P}\; = {\text{ }}n{\text{ }}{C_P}\Delta T$
This value is equal to the change in enthalpy, that is,
${q_P}\; = {\text{ }}n{\text{ }}{C_P}\Delta T{\text{ }} = {\text{ }}\Delta H$---------------[1]
Similarly, at constant volume V, we have
${q_V}\; = {\text{ }}n{\text{ }}{C_V}\Delta T$
This value of ${q_v}$, is equal to the change in internal energy of the system, that is,
${q_V}\; = {\text{ }}n{\text{ }}{C_V}\Delta T{\text{ }} = {\text{ }}\Delta U$---------------[2]
We know that for one mole of an ideal gas, when (n=1):

$$\Delta H{\text{ }} = {\text{ }}\Delta U{\text{ }} + {\text{ }}\Delta \left( {pV} \right){\text{ }}\left( {\because {\text{from ideal gas law PV = nRT = RT, for n = 1}}} \right)\; \\\ \Rightarrow \Delta H = {\text{ }}\Delta U{\text{ }} + {\text{ }}\Delta \left( {RT} \right){\text{ }} = {\text{ }}\Delta U{\text{ }} + {\text{ }}R{\text{ }}\Delta T \\\ \\\ {{\text{Therefore, }}\;\Delta H{\text{ }} = {\text{ }}\Delta U{\text{ }} + {\text{ }}R{\text{ }}\Delta T}$$

Substituting the values of ${\text{\Delta H and \Delta U}}$ from above in the former equation,[1] and [2]

$$\Rightarrow {C_P}\Delta T{\text{ }} = {\text{ }}{C_V}\Delta T{\text{ }} + {\text{ }}R{\text{ }}\Delta T \\\ \Rightarrow {C_P}\; = {\text{ }}{C_V}\; + {\text{ }}R \\\ \therefore {C_{P\;}}-{\text{ }}{C_V}\; = {\text{ }}R \\\$$

Note: Usually in chemistry texts they write the first law as$\Delta U = Q + W$. It is the same law, just the difference is that W is defined as the work done on the system instead of work done by the system.