If the ratio of the specific hearts of stream is $1.33 =$ and $R = 83121 / K m o l e K$ find the molar heat capacities of stream at constant pressure and volume.

33.5 k J / k m o l e, 25.19 k J / k g K

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24.93 J / k m o l e, 33.24 J / m o l K

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18.82 k J / k, m o l e, 10.82 k J / k m o l e

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24.12 k J / k, m o l e, 16.12 k J / k m o l e

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Solution

The correct option is A $33.5 k J / k m o l e, 25.19 k J / k g K$
Given : $\frac{C_{p}}{C_{v}} = 1.33$
We know that $C_{p} - C_{v} = R$
So, $1.33 C_{v} - C_{v} = R$
$⟹ C_{v} = 3 R = 3 \times 8.3121 = 24.93 J / K - m o l$
Also, $C_{p} = C_{v} + R = 24.93 + 8.3121 = 33.24 J / K - m o l$

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=

=

0.3122 k j / k g K

R

=

=

39.95

0.3122

8.314

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