$C_{P} 29 J m o l^{- 1} K^{- 1}$ in the process PT = constant.

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Solution

The process is given as $P T = c o n s t a n t$
From ideal gas equation $T = \frac{P V}{n R}$
$∴$ $P \times \frac{P V}{n R} = c o n s t a n t$

$⟹$ $P^{2} V = c o n s t a n t$ OR $P V^{1 / 2} = c o n s t a n t$
Comparing with $P V^{m} = c o n s t a n t$ , we get $m = \frac{1}{2}$

Using $C_{p} = \frac{R}{γ - 1} + \frac{R}{1 - m}$

$29 = \frac{8.314}{γ - 1} + \frac{8.314}{1 - \frac{1}{2}}$ $⟹$ $γ = 1.67$

$γ = 1 + \frac{2}{f}$ where $f$ is the degree of freedom
$∴$ $1.67 = 1 + \frac{2}{f}$ $⟹ f = 3$

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C = 29 J m o l^{- 1} K^{- 1}

C = 37.35 J {mol}^{- 1} K^{- 1}

P T = constant

(R = 2 c a l m o l^{- 1} K^{- 1})

C = 37.55

m o l e^{- 1} K^{- 1}

=

S_{C_{2} H_{6}}^{\circ} = 225 J m o l^{- 1} K^{- 1}, S_{C_{2} H_{4}}^{c i r c} = 220 J m o l^{- 1} K^{- 1}, S_{H_{2}}^{\circ} = 130 J m o l^{- 1} K^{- 1}

△ S^{\circ}

C_{2} H_{4} + H_{2} \to C_{2} H_{6}

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