Example 9.50b: Backward fading part 2

How many grams of gas are present in 8.75 L of \(\ce{C2H4}\) (ethene), at 378.3 kPa and 483 K?

\(V_{\mathrm{\ce{C2H4}}}\) \(= 0.100\ \mathrm{L}\)


\(P\) \(= 378.3\ \mathrm{kPa}\)

\(\ \ \ =3.734\ \mathrm{atm}\)


\(T\) \(= 483.\ \mathrm{K}\)


First, use the ideal gas law to obtain the chemical amount of ethene, \(n_{\mathrm{\ce{C2H4}}}\)
\(R\) \(= 8.314\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)

\(\ \ \ =0.08205\ \frac{\mathrm{L}\ \mathrm{atm}}{\mathrm{mol}\ \mathrm{K}}\)


\(n_{\mathrm{\ce{C2H4}}}\) \(= \dfrac{P \cdot V_{\mathrm{\ce{C2H4}}}}{R \cdot T}\)

\(\ \ \ =\dfrac{3.734\ \mathrm{atm} \cdot 0.100\ \mathrm{L}}{0.08205\ \frac{\mathrm{L}\ \mathrm{atm}}{\mathrm{mol}\ \mathrm{K}} \cdot 483.\ \mathrm{K}}\)

\(\ \ \ =\dfrac{0.3734\ \mathrm{L}\ \mathrm{atm}}{39.632\ \frac{\mathrm{L}\ \mathrm{atm}}{\mathrm{mol}}}\)

\(\ \ \ =9.4\times 10^{-3}\ \mathrm{mol}\)


a) Tell me about the first step. Why are we doing this?
b) Complete the second step, calculation of the mass of ethene, \(m_{\mathrm{\ce{C2H4}}}\)
\(m_{\mathrm{\ce{C2H4}}}\) = ?



Think about it...