The specific heat of Argon at constant volume is $0.3122$ kJ/kg K. Find the specific heat of Argon at constant pressure if R= $8.314$ kJ/ kmole K.

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Solution

$C_{p m} = \frac{5}{2} (R) = \frac{5}{2} (8.314) k J / k g m o l - K$

$C_{p m} = 20.785 k J / k g m o l - R$

$C_{v m} = C p m - R = \frac{3}{2} (R) = \frac{3}{2} (8.314) k J / k g m o l - K$

$C_{v m} = 12.471 k J / k g m o l - K$

$F o r A r g o n :$ M=39.95kg/kgmol$

$C_{v} = \frac{C_{v m}}{M}$

$C_{v} = \frac{12.471 k J / k g m o l - K}{39.95 k g / k g m o l} = 0.3122 k J / k g - K$

$C_{v} = (0.3122 k J / k g - K) * (1000 J / k J)$

$C_{v} = 312.2 J / k g - K$

$C_{p} = ⎡ ⎢ ⎢ ⎢ ⎣ \frac{\frac{5}{2}}{\frac{3}{2}} ⎤ ⎥ ⎥ ⎥ ⎦ C_{v}$

$C_{p} = \frac{5}{3} C_{v}$

$C_{p} = \frac{5}{3} [312.2 J / k g - K]$

$C_{p} = 520.3 J / k g - K$

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Similar questions

0.3122 k j / k g K

R

=

=

39.95

0.075 kcal/kg K

R = 2 cal/mol K

=

=

0.075 k c a l / k g - K

C_{v} = 2.98 c a l / m o l / K

= 6.02 \times 10^{23}

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