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Question: For hydrogen gas, \({{C}_{p}}\)= 3.409\(cal\,{{g}^{-1}}^{\,\circ }{{C}^{-1}}\), \({{C}_{v}}\)= 2.409...

For hydrogen gas, Cp{{C}_{p}}= 3.409calg1C1cal\,{{g}^{-1}}^{\,\circ }{{C}^{-1}}, Cv{{C}_{v}}= 2.409 calg1C1cal\,{{g}^{-1}}^{\,\circ }{{C}^{-1}} and molecular weight of hydrogen=2. Calculate the value of J.

Explanation

Solution

Before solving this question, we should first know what specific heat is. Specific heat is the amount of heat required to change the heat content of 1 gram of material by 1C^{\circ }C. The S.I unit of it is J/k/kg.

Complete answer:
The way to determine specific heat value is when two materials are kept in contact with each other and both of them initially is at a different temperature, heat always flows from the material which is warmer to the material which is colder until both the materials attain the same temperature. As the law of conservation of energy states the heat gained by the initially colder material should be equal to the heat lost by the initially warmer material. Molar heat capacity= Specific heat ×\times molar mass
Cp{{C}_{p}}= 3.409×\times 2 = 6.818 calmol1C1cal\,mo{{l}^{-1}}{{\,}^{\circ }}{{C}^{-1}}
Cv{{C}_{v}}= 2.409×\times 2 = 4.818 calmol1C1cal\,mo{{l}^{-1}}{{\,}^{\circ }}{{C}^{-1}}
When Cp{{C}_{p}}andCv{{C}_{v}} are in calmol1C1cal\,mo{{l}^{-1}}{{\,}^{\circ }}{{C}^{-1}}, when two gases are constant then the value of R should also be in the same units, and this can happen when we divide it by joule’s mechanical equivalent of heat i.e J.
Now, we have to use Meyer’s relation :
CpCv=RJ{{C}_{p}}-{{C}_{v}}=\dfrac{R}{J}
6.818 – 4.818 = 8.31J\dfrac{8.31}{J}
2 = 8.31J\dfrac{8.31}{J}
J = 4.16 J/ Cal

Note:
Heat is known as thermal energy. It is a form of energy that can be transformed from one to another (For example- A blender transforms electrical energy into mechanical energy) but it can neither be created nor be destroyed, energy is conserved in this. In thermodynamics, the higher the temperature of a material, the higher would be the thermal energy it possesses. Also, the more of a given substance at a given temperature, the more the material will possess total thermal energy.