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Problem

I want to compute relative error for volume when i have measured following data. $$ \left| \begin{matrix} \hline \text{Measured data}\\ \hline T=301.4\pm 0.1 \text{ K}\\ p=253.1\pm0.04 \text{ kPa} \\ \hline \end{matrix} \right| \quad \left| \begin{matrix} \hline \text{Constants} \\ \hline n=2.0 \text{ mol} \\ R=8.31447 \frac{\text{J}}{\text{K mol}} \\ \hline \end{matrix} \right| $$ $$ \left| \begin{matrix} \hline n=\text{amount of substance in gas(in moles)} \\ T=\text{The absolute temperature of gas} \\ p=\text{pressure of the gas} \\ V=\text{volume of the gas} \\ R=\text{gas constant} \\ \hline \end{matrix} \right| $$ wikipedia - Ideal gas law

Attempt to solve:

A expression for volume in this case can be easily derived from ideal gas law

$$ pV=nRT $$ Expression for volume is: $$ V=\frac{nRT}{p} $$

To my understanding we should be able to compute the relative error $\Delta V$ with partial differential equation:

$$ \Delta V = |\frac{\delta V}{\delta p}|\Delta p+ |\frac{\delta V}{\delta T}|\Delta T $$

$$ \Delta V =|-\frac{nRT}{p^2}|\Delta p + |\frac{nR}{p}|\Delta T$$

If we plug in the values

$$ \Delta V =|-\frac{2 \text{ mol}\cdot 8.31447 \frac{\text{J}}{\text{K mol}}\cdot 301.4 \text{ K}}{(253.1 \text{ kPa})^2}|\cdot0.04 + |\frac{2.0 \text{ mol}\cdot 8.31447 \frac{\text{J}}{\text{K mol}}}{253.1 \text{ kPa}}|\cdot0.1$$

$$ \Delta V \approx 9.699668356 \cdot 10^{-3} \text{ m}^3 $$ and in liters $$ \Delta V \approx 9.699668356 \text{ L} $$

$$ \Delta V \approx 9.7 \text{ L} $$


For some unknown reason this isn't the correct answer to this problem. If someone can spot the error that would be highly appreciated.

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I get by 10^-6, not 10^-3 in Mathcad:

enter image description here

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