Question

If the ratio of the specific heats of steam is 1.33 and R=8312 J/k mol K, find the molar heat capacities of steam at constant pressure and constant volume.
A.33.5 kJ/k mole, 25.19 kJ/k mole
B.25.19 kJ/k, 33.5 kJ/kg K
C.18.82 kJ/k, 10.82 kJ/mol K
D.None of these

Answer 1

Hint: In this question we have to find the specific heat capacities at constant volume and pressure. For this we are going to use Mayer’s formula and we will use the formula of the ratio of specific heats at constant volume and pressure.

Complete step by step answer:
Given,
\[\dfrac{{{C_P}}}{{{C_V}}} = 1.33\]
R=8312 J/k mole
 Formulae used,
\[\dfrac{{{C_P}}}{{{C_V}}} = \gamma \]….. (1)
\[{C_P} - {C_V} = R\]……. (2)
If we convert above formula to find the relation between\[{C_V}\], R and \[\gamma \]
\[{C_V} = \dfrac{R}{{\gamma - 1}}\]…. (3)
Where,
\[{C_P}\] is specific heat at constant pressure and
\[{C_V}\] is specific heat at constant volume
Now, using given values to find \[{C_P}\] and \[{C_V}\]
R is gas constant
\[{C_V} = \dfrac{R}{{\gamma - 1}}\]
\[{C_V} = \dfrac{{8312}}{{1.33 - 1}}\]
So on doing the simplification,we have
\[{C_V} = \dfrac{{8312}}{{0.33}}\]
\[{C_V} = 25187.87\] J/k mole
\[{C_V} = 25.19\] kJ/k mole
Now, using equation (2) to find\[{C_P}\], putting the value of \[{C_V}\]in equation (2)
\[{C_P} = R + {C_V}\]
\[{C_P} = 8312 + 25187.87\]
And hence on solving, we have
\[{C_P} = 33499.87\] J/k mole
\[{C_P} = 33.5\] kJ/k mole
Result- Hence, from the above calculation the specific heat at constant volume is \[{C_V} = 25.19\] kJ/k mol and \[{C_P} = 33.5\] kJ/k mole.
Hence option (A) is correct.

Note:
In this question we have used two formulae of specific heat. So, the knowledge of these formulae is a must and we should be careful while doing the calculation of the problems. One more thing that we must be careful about is the units of parameters. All the units must be in the same system.
In this question it was quite easy that we just had to apply formula directly but there might be a case when we have to be careful when using the formulae because we might have to use some theoretical aspects before applying the relevant formula. By following these steps we can find the correct answer to the problem.