Question

Complete the following table for dinitrogen tetroxide gas. \begin{tabular}{llcll} \hline \multicolumn{1}{c} { Pressure } & Volume & Temperature & Moles & Grams \\\ \hline (a) \(1.77 \mathrm{~atm}\) & \(4.98 \mathrm{~L}\) & \(43.1^{\circ} \mathrm{C}\) & & \\ \cline { 4 - 5 } (b) \(673 \mathrm{~mm} \mathrm{Hg}\) & \(488 \mathrm{~mL}\) & & 0.783 & \\ \hline (c) \(0.899 \mathrm{bar}\) & & \(912^{\circ} \mathrm{C}\) & & 6.25 \\ (d) & \(1.15 \mathrm{~L}\) & \(39^{\circ} \mathrm{F}\) & 0.166 & \\ \hline \end{tabular}

Short answer

Based on the Ideal Gas Law, find the missing values for each of the following cases of dinitrogen tetroxide gas: a. P = 1.77 atm, V = 4.98 L, T = 43.1 °C. b. P = 673 mm Hg, V = 488 mL, n = 0.783 moles. c. P = 0.899 bar, T = 912 °C, 6.25 g. d. V = 1.15 L, T = 39 °F, n = 0.166 moles. Please provide your solution in grams for a, in Celsius for b, in liters for c, and in mm Hg for d.

Step-by-step solution

Set up the Ideal Gas Law for each case

For each case (a), (b), (c), and (d), we will set up the equation PV = nRT with the given values. If needed, we will also convert the units to match the equation requirements. #a.# Pressure = 1.77 atm (already in atm) Volume = 4.98 L (already in L) Temperature = \(43.1+273.15 = 316.25 K\) (converted to Kelvin) #b.# Pressure = \(673\,\text{mm}\,\text{Hg} *\frac{1\,\text{atm}}{760\,\text{mm}\,\text{Hg}}=0.8855\,\text{atm}\) (converted to atm) Volume = \(488 \,\text{mL} \ *\frac{1\,\text{L}}{1000\,\text{mL}} = 0.488\,\text{L}\) (converted to L) Moles = 0.783 moles (given) #c.# Pressure = \(0.899\,\text{bar}\ *\frac{1\,\text{atm}}{1.01325\,\text{bar}}= 0.8876\,\text{atm}\) (converted to atm) Temperature = \(912+273.15 = 1185.15 K\) (converted to Kelvin) Grams = 6.25 g (given) #d.# Volume = 1.15 L (already in L) Temperature = \(((39-32)* \frac{5}{9})+273.15 = 277.04 K\) (converted to Fahrenheit to Kelvin) Moles = 0.166 moles (given) We can now use the Ideal Gas Law to find the missing values in each case.

Finding Missing Values#a. Calculating moles (n)

Using PV = nRT, solving for n: \(n = \frac{PV}{RT} = \frac{(1.77)(4.98)}{(0.0821)(316.25)} =0.1657\,\text{moles} \) Now, to find the grams. The molecular weight of dinitrogen tetroxide (N2O4) = 2 * 14.01 (nitrogen) + 4 * 16.00 (oxygen) = 92.02 g/mol. Grams = moles * molecular weight \(= 0.1657 * 92.02 = 15.24\,\text{g}\)

b. Calculating temperature (T)

Using PV = nRT, solving for T: \(T=\frac{PV}{nR} = \frac{(0.8855)(0.488)}{(0.0821)(0.783)} = 324.40\,\text{K}\) Now, convert temperature back to Celsius: \(°\text{C} = 324.40 - 273.15 = 51.25^\circ\,\text{C}\)

c. Calculating Volume (V)

Using PV = nRT, solving for V: First, find the moles (n): Moles = \(\frac{\text{grams}}{\text{molecular weight}}\) \(n = \frac{6.25}{92.02} = 0.0679\,\text{moles}\) Now, find the volume: \(V=\frac{nRT}{P} = \frac{(0.0679)(0.0821)(1185.15)}{0.8876} = 7.3064\,\text{L}\)

d. Calculating pressure (P)

Using PV = nRT, solving for P: \(P = \frac{nRT}{V} = \frac{(0.166)(0.0821)(277.04)}{1.15} = 3.019\,\text{atm}\) Now, convert the pressure back to mm Hg: \(3.019\,\text{atm}* \frac{760\,\text{mm}\,\text{Hg}}{1\,\text{atm}} = 2294\,\text{mm}\,\text{Hg}\)