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Say you are given the equation

$$ p(V-b) = RTe^{-\frac{a}{RVT}} $$

Where p, b, R and a are all constants.

And you are asked to find an expression for $\frac{dV}{dT}$, would the left side be a chain rule with a product rule with an implicit differential?

So far I believe that the right hand side differentiates to:

$$ \frac{a}{TV}e^{\frac{-a}{RVT}}+\frac{dV}{dT}*\frac{a}{V^2}e^{\frac{-1}{RVT}} $$

Is that fully correct?

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We have

$$p(V-b) = RT e^{-\frac{a}{RVT}}\\ \implies pdV=\left(Re^{-\frac{a}{RVT}}+RTe^{-\frac{a}{RVT}}\left(\frac{a}{RVT^2}\right)\right)dT+RTe^{-\frac{a}{RVT}}\left(\frac{a}{RV^2T}\right)dV=\\ =\left(Re^{-\frac{a}{RVT}}+e^{-\frac{a}{RVT}}\left(\frac{a}{VT}\right)\right)dT+e^{-\frac{a}{RVT}}\left(\frac{a}{V^2}\right)dV$$

then

$$\frac{dV}{dT} =\frac{Re^{-\frac{a}{RVT}}+\frac{a}{VT}e^{-\frac{a}{RVT}}}{p-\frac{a}{V^2}e^{-\frac{a}{RVT}}} =\frac{Re^{-\frac{a}{RVT}}+\frac{a}{VT}e^{-\frac{a}{RVT}}}{\frac{RT e^{-\frac{a}{RVT}}}{V-b}-\frac{a}{V^2}e^{-\frac{a}{RVT}}} =\frac{R+\frac{a}{VT}}{\frac{RT }{V-b}-\frac{a}{V^2}} = \\=\frac{(RVT+a)(V-b)V^2}{(RTV^2-a(V-b))VT } =\frac{(RVT+a)(V-b)V}{RT^2V^2-aT(V-b)}$$

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Take the log on both sides

$$ \log p + \log(V-b)=\log R + \log T - \frac{a}{R V T} $$

and then

$$ \frac{dV}{V-b} = \frac{dT}{T}+\frac{a dV}{R T V^2}+\frac{a dT}{R T^2 V} $$

etc.

giving

$$ \frac{dV}{dT} = \frac{V (V-b) (a+R T V)}{T \left(a (b-V)+R T V^2\right)} $$

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$$\ln[p(V-b)]=\ln[RTe^{{-\frac{a}{RVT}}}]$$ $$\ln(p)+\ln(V-b)=\ln(R)+\ln(T)-\frac{a}{RVT}$$ $$\frac{d}{dT}\left[\ln(p)+\ln(V-b)=\ln(R)+\ln(T)-\frac{a}{RVT}\right]$$ $$\frac{\frac{dV}{dT}}{V-b}=\frac{1}{T}-\frac{a}{R}\left[\frac{d}{dT}\left(\frac{1}{VT}\right)\right]\qquad \text{Apply Quotient Rule}$$ $$\frac{\frac{dV}{dT}}{V-b}=\frac{1}{T}+\frac{a}{R}\left[\frac{d}{dT}(VT)\over(VT)^2\right]$$ $$\frac{\frac{dV}{dT}}{V-b}=\frac{1}{T}+\frac{a}{R}\left[V+T\frac{dV}{dT}\over(VT)^2\right]$$ $$\frac{dV}{dT}=\frac{V-b}{T}+\frac{a(V-b)}{R}\left[V+T\frac{dV}{dT}\over(VT)^2\right]$$ $$\frac{dV}{dT}=\frac{V-b}{T}+\frac{a(V-b)}{RVT^2}+\left[\frac{a(V-b)}{RV^2T}\right]\frac{dV}{dT}$$ $$\frac{dV}{dT}=\left[\frac{RV^2T(V-b)}{RV^2T-a(V-b)}\right]\left[\frac{1}{T}+\frac{a}{RVT^2}\right]$$ $$\frac{dV}{dT}=\left[\frac{RV^2T(V-b)}{RV^2T-a(V-b)}\right]\left[\frac{RVT^2+aT}{RVT^3}\right]$$ $$\frac{dV}{dT}=\left[\frac{V(V-b)}{RV^2T-a(V-b)}\right]\left[\frac{RVT^2+aT}{T^2}\right]$$ $$\color{red}{\mathbf{\frac{dV}{dT}=\frac{V(V-b)(a+RVT)}{T\left[RV^2T-a(V-b)\right]}}}$$

And That's all, This is the answer

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  • $\begingroup$ It seems there is an error here $$\frac{dV}{dT}=\left[\frac{RV^2T(V-b)}{RV^2T-a(V+b)}\right]\left[\frac{RVT^2+aT}{RVT^3}\right] \implies \frac{dV}{dT}=\left[\frac{V(V-b)}{RV^2T-a(V+b)}\right]\left[\frac{RVT+a}{T}\right] $$ $\endgroup$ – gimusi May 15 '18 at 13:25
  • $\begingroup$ Thanks, I fixed it :) $\endgroup$ – MR ASSASSINS117 May 15 '18 at 14:28
  • $\begingroup$ Maybe we need an extra factor T at the denominator! $\endgroup$ – gimusi May 15 '18 at 14:29

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