Question 741: Thermodynamics & KTG

Question

Four moles of hydrogen, two moles of helium and one mole of water vapour form an ideal gas mixture. $\left[C_{\mathrm{v}}\right.$ for hydrogen $=\frac{5}{2} R, C_v$ for helium $=\frac{3}{2} R, \quad C_{\mathrm{v}}$ for water vapour $\left.=3 \mathrm{R}\right]$ What is the molar specific heat at constant pressure of the mixture?

Accepted Answer

To find the molar specific heat at constant pressure for the mixture, you can use the formula for molar specific heat at constant pressure, $C_p$, which is given by:
$ C_p = C_v + R $
Using this formula, calculate $C_p$ for each component of the gas mixture:
For hydrogen, the molar specific heat at constant volume is $C_v = \frac{5}{2} R$, so:
$   C_{p1} = \frac{5}{2} R + R = \frac{7}{2} R   $
For helium, $C_v = \frac{3}{2} R$, so:
$   C_{p2} = \frac{3}{2} R + R = \frac{5}{2} R   $
For water vapor, $C_v = 3R$, so:
$   C_{p3} = 3R + R = 4R   $
The number of moles for each gas is given:
$n_1 = 4$ moles of hydrogen
$n_2 = 2$ moles of helium
$n_3 = 1$ mole of water vapor
The total number of moles, $n_{	ext{total}}$, is:
$ n_{	ext{total}} = n_1 + n_2 + n_3 = 4 + 2 + 1 = 7 $
Now, calculate the molar specific heat at constant pressure for the mixture using the weighted average of the specific heats:
$ C_p 	ext{ of mixture} = \frac{n_1 C_{p1} + n_2 C_{p2} + n_3 C_{p3}}{n_{	ext{total}}} $
Substitute the values:
$ C_p 	ext{ of mixture} = \frac{4 	imes \frac{7}{2} R + 2 	imes \frac{5}{2} R + 1 	imes 4R}{7} $
Perform the calculations:
$ = \frac{14R + 5R + 4R}{7} $
$ = \frac{23R}{7} $
Therefore, the molar specific heat of the mixture at constant pressure is $\frac{23}{7} R$.

Answer

$\frac{11}{3} \mathrm{R}$

Answer

$\frac{23}{7} R$

Answer

$\frac{16}{7} R$

Answer

$\frac{23}{3} R$

Thermodynamics & KTG

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