Question

thermodyanamics numericals
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Answer 1

Answer:

if

the gas has n1 moles, then the amount of heat energy Q1 transferred to a body having heat capacity C1 will be,

Q1 = n1C1 ΔT

Similarly, if the gas has n2 moles, then the amount of heat energy Q2 transferred to a body having heat capacity C2 will be,

Q2 = n2C2 ΔT

And

if the gas has n3 moles, then the amount of heat energy Q3 transferred to a body having heat capacity C3 will be,

Q3 = n3C3 ΔT

As, each species will experience the same temperature change, thus,

Q = Q1 + Q2 + Q3

   = n1C1 ΔT + n2C2 ΔT + n3C3 ΔT

Dividing both the sides by n = n1 + n2 + n3 and ΔT, then we will get,

Q/nΔT = (n1C1 ΔT + n2C2 ΔT + n3C3 ΔT)/ nΔT

As, Q/nΔT = C, thus,

          C=ΔT (n1C1 + n2C2  + n3C3)/ nΔT

          = (n1C1 + n2C2  + n3C3)/ n

          = n1C1 + n2C2  + n3C3/ n1 + n2 + n3

From the above observation we conclude that, the molar specific heat at constant volume of the mixture would be n1C1 + n2C2  + n3C3/ n1 + n2 + n3.