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For an ideal ...

Question

For an ideal gas, the slope of $V - T$ graph during an adiabatic process is $\frac{d V}{d T} = - m$ at a point where volume and temperature are $V_{0}$ and $T_{0}$ . Find the value of $C_{P}$ of the gas. It is given that $m$ is a positive number.

A

(\frac{m T_{0}}{V_{0}} + 1) R

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B

\frac{m T_{0} R}{V_{0}}

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C

(\frac{- m T_{0}}{V_{0}} + 1) R

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D

(\frac{- m T_{0} R}{V_{0}})

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Solution

The correct option is A $(\frac{m T_{0}}{V_{0}} + 1) R$
For an adiabatic process, the equation of state can be written as
$T V^{γ - 1} =$ constant
Applying logarithm on both sides,
$ln T + (γ - 1) ln V =$ constant
Differentiating both sides with respect to $^{'} T^{'}$ , we get
$\frac{1}{T} + \frac{γ - 1}{V} \frac{d V}{d T} = 0$
$\Rightarrow \frac{d V}{d T} = - \frac{V}{T (γ - 1)} . . . . . . (1)$
Given that, ${(\frac{d V}{d T})}_{V_{0}, T_{0}} = - m$
$∴$ From $(1)$ , we can write that $\frac{V_{0}}{T_{0} (γ - 1)} = m \Rightarrow \frac{1}{γ - 1} = \frac{m T_{0}}{V_{0}}$
Now, $C_{V} = \frac{R}{γ - 1} = \frac{m R T_{0}}{V_{0}}$
$& C_{p} = C_{V} + R = (\frac{m T_{0}}{V_{0}} + 1) R$
Thus, option (a) is the correct answer.

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