Answer to Question #111081 in General Chemistry for jay

Question #111081

The pressure of a mixture of nitrogen, carbon dioxide and oxygen is 195 kPa. What is the partial pressure of oxygen if the partial pressures of the nitrogen and carbon dioxide are 80 kPa and 55 kPa, respectively? If a gas has an initial pressure of 1216 kPa, a volume of 23.0 L, and a temperature of 200 K and the pressure increases to 1418 kPa while the temperature increases to 300 K, what is the new volume of the gas? If a container of gas has a temperature of 222 K and a pressure of 167 kPa, to what temperature would you have to heat the container to achieve a pressure of 520 kPa? How many moles of gas are present in a 562 mL container if the pressure in the container is 340 kPa and the container if 87 degrees Celsius? If a gas has a volume of 2.52 L and a pressure of 180 kPa, what will the pressure be if the volume is changed to 3.15 L?

Expert's answer

1) The overall pressue P of the mixture of gases is a sum of their partial pressures p_i. Then

p(O₂) = P - p(N₂) - p(CO₂) = 195 - 80 - 55 = 60 kPa

2) According to Ideal gas law,

PV = nRT. Number of moles remains unchanged, but other parameters change.

"\\begin{cases}\n P_1V_1=nRT_1\\\\\n P_2V_2 = nRT_2\n\\end{cases}\n\\begin{cases}\n nR={\\frac {P_1V_1}{T_1}}\\\\\n nR={\\frac {P_2V_2}{T_2}}\n\\end{cases}\\\\\n {\\frac {P_1V_1}{T_1}}={\\frac {P_2V_2}{T_2}}\\\\\nV_2=V_1{\\frac {P_1T_2}{P_2T_1}}=23*{\\frac {1216*300}{1418*200}}=29.6 L"

3) There, not only number of moles remains unchanged, but volume as well.

"\\begin{cases}\n P_1V=nRT_1\\\\\n P_2V = nRT_2\n\\end{cases}\n\\begin{cases}\n {\\frac {nR}{V}}={\\frac {P_1}{T_1}}\\\\\n {\\frac {nR}{V}}={\\frac {P_2}{T_2}}\n\\end{cases}\\\\\n {\\frac {P_1}{T_1}}={\\frac {P_2}{T_2}}\\\\\nT_2=T_1{\\frac {P_2}{P_1}}=222*{\\frac {520}{167}}=691 K"