4 added 105 characters in body edited Jan 25 at 15:14 Buck Thorn 5,98822 gold badges88 silver badges3434 bronze badges I'd start the solution with this equation: $$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$$ where I emphasize that the density of the gas is a molar average value, and rewrite it as $$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$$ where $$\chi_{Ar} = n_{Ar}/n$$ is the mole fraction of argon in the gas mixture and the $$d_{i}$$ are densities computed assuming all of the gas corresponds to He or Ar, for instance $$d_{Ar} = M_{Ar}n/V = M_{Ar}P/RT$$$$d_{Ar} = M_{Ar}n/V = M_{Ar}/V_{m}$$ I add a mea culpa here, as this was massively botched in a previous answer but stands corrected nowwhere $$V_{m}$$ is the molar volume at STP (22.414 $$m^3/kg mol$$). It follows that $$d_{avg} = \chi_{Ar}M_{Ar}P/RT + (1-\chi_{Ar})M_{He}P/RT = (\chi_{Ar}M_{Ar} + (1-\chi_{Ar})M_{He})P/RT$$$$d_{avg} = \chi_{Ar}M_{Ar}/V_{m} + (1-\chi_{Ar})M_{He}/V_{m} = (\chi_{Ar}M_{Ar} + (1-\chi_{Ar})M_{He})/V_{m} = M_{avg}/V_{m}$$ Thiswhich can be rearranged assolved for $$\chi_{Ar}$$: $$\chi_{Ar} = ((d_{avg}RT/P)-M_{He})/(M_{Ar}-M_{He})$$$$\chi_{Ar} = ((d_{avg}V_{m})-M_{He})/(M_{Ar}-M_{He})$$ Finally, the molar percentage argon is computed as $$f_{Ar}=100\chi_{Ar}$$. For your particular problem I get $$f_{Ar}$$= 5045.14%00%. I'd start the solution with this equation: $$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$$ where I emphasize that the density of the gas is a molar average value, and rewrite it as $$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$$ where $$\chi_{Ar} = n_{Ar}/n$$ is the mole fraction of argon in the gas mixture and the $$d_{i}$$ are densities computed assuming all of the gas corresponds to He or Ar, for instance $$d_{Ar} = M_{Ar}n/V = M_{Ar}P/RT$$ I add a mea culpa here, as this was massively botched in a previous answer but stands corrected now. It follows that $$d_{avg} = \chi_{Ar}M_{Ar}P/RT + (1-\chi_{Ar})M_{He}P/RT = (\chi_{Ar}M_{Ar} + (1-\chi_{Ar})M_{He})P/RT$$ This can be rearranged as $$\chi_{Ar} = ((d_{avg}RT/P)-M_{He})/(M_{Ar}-M_{He})$$ Finally, the molar percentage argon is computed as $$f_{Ar}=100\chi_{Ar}$$. For your particular problem I get $$f_{Ar}$$= 50.14%. I'd start the solution with this equation: $$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$$ where I emphasize that the density of the gas is a molar average value, and rewrite it as $$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$$ where $$\chi_{Ar} = n_{Ar}/n$$ is the mole fraction of argon in the gas mixture and the $$d_{i}$$ are densities computed assuming all of the gas corresponds to He or Ar, for instance $$d_{Ar} = M_{Ar}n/V = M_{Ar}/V_{m}$$ where $$V_{m}$$ is the molar volume at STP (22.414 $$m^3/kg mol$$). It follows that $$d_{avg} = \chi_{Ar}M_{Ar}/V_{m} + (1-\chi_{Ar})M_{He}/V_{m} = (\chi_{Ar}M_{Ar} + (1-\chi_{Ar})M_{He})/V_{m} = M_{avg}/V_{m}$$ which can be solved for $$\chi_{Ar}$$: $$\chi_{Ar} = ((d_{avg}V_{m})-M_{He})/(M_{Ar}-M_{He})$$ Finally, the molar percentage argon is computed as $$f_{Ar}=100\chi_{Ar}$$. For your particular problem I get $$f_{Ar}$$= 45.00%. Post Undeleted by Buck Thorn occurred Jan 25 at 14:20 3 deleted 253 characters in body edited Jan 25 at 14:15 Buck Thorn 5,98822 gold badges88 silver badges3434 bronze badges I'd start the solution with this equation: $$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$$ where I emphasize that the density of the gas is a molar average value, and rewrite it as $$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$$ where $$\chi_{Ar} = n_{Ar}/n$$ is the mole fraction of argon in the gas mixture. Insert and the expression $$d_{Ar} = M_{Ar}n_{Ar}/V = M_{Ar}\chi_{Ar}n/V= M_{Ar}\chi_{Ar}P/RT$$ for the density$$d_{i}$$ are densities computed assuming all of argon in the mixture ($$P\chi_{Ar}=P_{Ar}$$ is the partial pressure of argon)gas corresponds to He or Ar, and the similar expression for helium, so that instance $$\chi_{Ar} \chi_{Ar}M_{Ar}P/RT + (1-\chi_{Ar})(1-\chi_{Ar})M_{He}P/RT = d_{avg}$$$$d_{Ar} = M_{Ar}n/V = M_{Ar}P/RT$$ Rearranging the equationI add a mea culpa here, $$\chi_{Ar}^2(M_{Ar}+M_{He}) - 2M_{He}\chi_{Ar} +(M_{He}- d_{avg}(RT/P))=0$$ or $$\chi_{Ar}^2 +\alpha\chi_{Ar} +\beta=0$$ where as this was massively botched in a previous answer but stands corrected now. $$\alpha = - 2M_{He}/(M_{Ar}+M_{He})$$ It follows that $$\beta = (M_{He}- (d_{avg}RT/P))/(M_{Ar}+M_{He})$$$$d_{avg} = \chi_{Ar}M_{Ar}P/RT + (1-\chi_{Ar})M_{He}P/RT = (\chi_{Ar}M_{Ar} + (1-\chi_{Ar})M_{He})P/RT$$ Finally, solving the quadratic equation forThis can be rearranged as $$\chi_{Ar}$$: $$\chi_{Ar} = -alpha/2+(1/2)(alpha^2-4\beta)^{1/2}$$$$\chi_{Ar} = ((d_{avg}RT/P)-M_{He})/(M_{Ar}-M_{He})$$ Finally, the molar percentage argon is computed as $$f_{Ar}=100\chi_{Ar}$$. For your particular problem I get $$f_{Ar}$$= 7350.79%14%. I'd start the solution with this equation: $$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$$ where I emphasize that the density of the gas is a molar average value, and rewrite it as $$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$$ where $$\chi_{Ar} = n_{Ar}/n$$ is the mole fraction of argon in the gas mixture. Insert the expression $$d_{Ar} = M_{Ar}n_{Ar}/V = M_{Ar}\chi_{Ar}n/V= M_{Ar}\chi_{Ar}P/RT$$ for the density of argon in the mixture ($$P\chi_{Ar}=P_{Ar}$$ is the partial pressure of argon), and the similar expression for helium, so that $$\chi_{Ar} \chi_{Ar}M_{Ar}P/RT + (1-\chi_{Ar})(1-\chi_{Ar})M_{He}P/RT = d_{avg}$$ Rearranging the equation, $$\chi_{Ar}^2(M_{Ar}+M_{He}) - 2M_{He}\chi_{Ar} +(M_{He}- d_{avg}(RT/P))=0$$ or $$\chi_{Ar}^2 +\alpha\chi_{Ar} +\beta=0$$ where $$\alpha = - 2M_{He}/(M_{Ar}+M_{He})$$ $$\beta = (M_{He}- (d_{avg}RT/P))/(M_{Ar}+M_{He})$$ Finally, solving the quadratic equation for $$\chi_{Ar}$$: $$\chi_{Ar} = -alpha/2+(1/2)(alpha^2-4\beta)^{1/2}$$ Finally, the molar percentage argon is computed as $$f_{Ar}=100\chi_{Ar}$$. For your particular problem I get $$f_{Ar}$$= 73.79%. I'd start the solution with this equation: $$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$$ where I emphasize that the density of the gas is a molar average value, and rewrite it as $$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$$ where $$\chi_{Ar} = n_{Ar}/n$$ is the mole fraction of argon in the gas mixture and the $$d_{i}$$ are densities computed assuming all of the gas corresponds to He or Ar, for instance $$d_{Ar} = M_{Ar}n/V = M_{Ar}P/RT$$ I add a mea culpa here, as this was massively botched in a previous answer but stands corrected now. It follows that $$d_{avg} = \chi_{Ar}M_{Ar}P/RT + (1-\chi_{Ar})M_{He}P/RT = (\chi_{Ar}M_{Ar} + (1-\chi_{Ar})M_{He})P/RT$$ This can be rearranged as $$\chi_{Ar} = ((d_{avg}RT/P)-M_{He})/(M_{Ar}-M_{He})$$ Finally, the molar percentage argon is computed as $$f_{Ar}=100\chi_{Ar}$$. For your particular problem I get $$f_{Ar}$$= 50.14%. Post Deleted by Buck Thorn occurred Jan 25 at 13:40 Post Undeleted by Buck Thorn occurred Jan 25 at 10:43 2 added 7 characters in body edited Jan 25 at 10:43 Buck Thorn 5,98822 gold badges88 silver badges3434 bronze badges I'd start the solution with this equation: $$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$$ where I emphasize that the density of the gas is ana averagemolar average value, and rewrite it as $$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$$ where $$\chi_{Ar} = n_{Ar}/n$$ is the mole fraction of argon in the gas mixture. Insert the expression $$d_{Ar} = M_{Ar}n_{Ar}/V = M_{Ar}\chi_{Ar}n/V= M_{Ar}\chi_{Ar}P/RT$$ for the density of argon in the mixture ($$P\chi_{Ar}=P_{Ar}$$ is the partial pressure of argon), and the similar expression for helium, so that $$\chi_{Ar} \chi_{Ar}M_{Ar}P/RT + (1-\chi_{Ar})(1-\chi_{Ar})M_{He}P/RT = d_{avg}$$ Rearranging the equation, $$\chi_{Ar}^2(M_{Ar}+M_{He}) - 2M_{He}\chi_{Ar} +(M_{He}- d_{avg}(RT/P))=0$$ or $$\chi_{Ar}^2 +\alpha\chi_{Ar} +\beta=0$$ where $$\alpha = - 2M_{He}/(M_{Ar}+M_{He})$$ $$\beta = (M_{He}- (d_{avg}RT/P))/(M_{Ar}+M_{He})$$ Finally, solving the quadratic equation for $$\chi_{Ar}$$: $$\chi_{Ar} = -alpha+(1/2)(alpha^2-4\beta)^{1/2}$$$$\chi_{Ar} = -alpha/2+(1/2)(alpha^2-4\beta)^{1/2}$$ Finally, the molar percentage argon is computed as $$f_{Ar}=100\chi_{Ar}$$. For your particular problem I get $$f_{Ar}$$= 8273.89%79%. I'd start the solution with this equation: $$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$$ where I emphasize that the density of the gas is an average value, and rewrite it as $$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$$ where $$\chi_{Ar} = n_{Ar}/n$$ is the mole fraction of argon in the gas mixture. Insert the expression $$d_{Ar} = M_{Ar}n_{Ar}/V = M_{Ar}\chi_{Ar}n/V= M_{Ar}\chi_{Ar}P/RT$$ for the density of argon in the mixture ($$P\chi_{Ar}=P_{Ar}$$ is the partial pressure of argon), and the similar expression for helium, so that $$\chi_{Ar} \chi_{Ar}M_{Ar}P/RT + (1-\chi_{Ar})(1-\chi_{Ar})M_{He}P/RT = d_{avg}$$ Rearranging the equation, $$\chi_{Ar}^2(M_{Ar}+M_{He}) - 2M_{He}\chi_{Ar} +(M_{He}- d_{avg}(RT/P))=0$$ or $$\chi_{Ar}^2 +\alpha\chi_{Ar} +\beta=0$$ where $$\alpha = - 2M_{He}/(M_{Ar}+M_{He})$$ $$\beta = (M_{He}- (d_{avg}RT/P))/(M_{Ar}+M_{He})$$ Finally, solving the quadratic equation for $$\chi_{Ar}$$: $$\chi_{Ar} = -alpha+(1/2)(alpha^2-4\beta)^{1/2}$$ Finally, the molar percentage argon is computed as $$f_{Ar}=100\chi_{Ar}$$. For your particular problem I get $$f_{Ar}$$= 82.89%. I'd start the solution with this equation: $$n_{Ar}d_{Ar}/(n_{Ar} + n_{He}) + n_{He}d_{He}/(n_{Ar} + n_{He}) = d_{avg}$$ where I emphasize that the density of the gas is a molar average value, and rewrite it as $$\chi_{Ar} d_{Ar} + (1-\chi_{Ar})d_{He} = d_{avg}$$ where $$\chi_{Ar} = n_{Ar}/n$$ is the mole fraction of argon in the gas mixture. Insert the expression $$d_{Ar} = M_{Ar}n_{Ar}/V = M_{Ar}\chi_{Ar}n/V= M_{Ar}\chi_{Ar}P/RT$$ for the density of argon in the mixture ($$P\chi_{Ar}=P_{Ar}$$ is the partial pressure of argon), and the similar expression for helium, so that $$\chi_{Ar} \chi_{Ar}M_{Ar}P/RT + (1-\chi_{Ar})(1-\chi_{Ar})M_{He}P/RT = d_{avg}$$ Rearranging the equation, $$\chi_{Ar}^2(M_{Ar}+M_{He}) - 2M_{He}\chi_{Ar} +(M_{He}- d_{avg}(RT/P))=0$$ or $$\chi_{Ar}^2 +\alpha\chi_{Ar} +\beta=0$$ where $$\alpha = - 2M_{He}/(M_{Ar}+M_{He})$$ $$\beta = (M_{He}- (d_{avg}RT/P))/(M_{Ar}+M_{He})$$ Finally, solving the quadratic equation for $$\chi_{Ar}$$: $$\chi_{Ar} = -alpha/2+(1/2)(alpha^2-4\beta)^{1/2}$$ Finally, the molar percentage argon is computed as $$f_{Ar}=100\chi_{Ar}$$. For your particular problem I get $$f_{Ar}$$= 73.79%. Post Deleted by Buck Thorn occurred Jan 25 at 10:14 1 answered Jan 25 at 10:11 Buck Thorn 5,98822 gold badges88 silver badges3434 bronze badges