Question

Suppose two 200.0 -L tanks are to be filled separately with the gases helium and hydrogen. What mass of each gas is needed to produce a pressure of 135 atm in its respective tank at \(24^{\circ} \mathrm{C} ?\)

Short answer

To produce a pressure of 135 atm in two separate 200.0-L tanks at \(24^{\circ} \mathrm{C}\), 4421.88 g of helium and 2210.94 g of hydrogen are needed.

Step-by-step solution

Convert Temperature to Kelvin

To convert the temperature from Celsius to Kelvin, we add 273.15 to the Celsius value: \(T_{K} = T_{\mathrm{C}} + 273.15\) \(T_{K} = 24^{\circ} \mathrm{C} + 273.15 = 297.15\ \mathrm{K}\)

Calculate the Number of Moles for Each Gas

Using the Ideal Gas Law, \(PV = nRT\), we will find the number of moles (n) for helium and hydrogen in their respective tanks. Since both tanks have the same volume and pressure, the number of moles will be the same for both gases: \(n = \frac{PV}{RT}\) \(n = \frac{135\ \mathrm{atm} \times 200.0\ \mathrm{L}}{0.0821\ \mathrm{L\ atm/mol\ K} \times 297.15\ \mathrm{K}}\) \(n = 1105.47\ \mathrm{moles}\)

Calculate the Mass of Each Gas

To calculate the mass of each gas, we multiply the number of moles by their respective molar masses: For helium, the molar mass is 4 g/mol: \(m_{He} = n \times M_{He}\) \(m_{He} = 1105.47\ \mathrm{moles} \times 4\ \mathrm{g/mol} = 4421.88\ \mathrm{g}\) For hydrogen, the molar mass is 2 g/mol: \(m_{H2} = n \times M_{H2}\) \(m_{H2} = 1105.47\ \mathrm{moles} \times 2\ \mathrm{g/mol} = 2210.94\ \mathrm{g}\) So, to produce a pressure of 135 atm in each 200.0-L tank at \(24^{\circ} \mathrm{C}\): - 4421.88 g of helium is needed. - 2210.94 g of hydrogen is needed.