Assume that you have an open-end manometer filled with ethyl alcohol 1density = 0.7893 g>mL at 20 °C2 rather than mercury 1density = 13.546 g>mL at 20 °C2. What is the pressure in pascals if the level in the arm open to the atmosphere is 55.1 cm higher than the level in the arm connected to the gas sample and the atmospheric pressure reading is 752.3 mm Hg?

Video transcript

Hi everyone. This problem reads at 20 degrees Celsius glycerol is used in an open end manama. Ter instead of mercury, calculate the gas pressure in atmospheres. If the level of the arm to the gas is 62.5 centimeters higher than the level to the atmosphere. The atmospheric pressure is 0.942 ATM Here's okay. So our goal here is to calculate the gas pressure in atmospheres and our gas pressure is going to be less than our atmospheric pressure. So we can go ahead and write that. The pressure of the gas, which is what we're looking for, is less than 0. atmospheres. Okay. And so let's go ahead and write out what we're given. So for glycerol Were given the density and the density is 1.261 grams per mil leader and we're told that The gas is 62.5 cm higher than the level of the atmosphere. So that means this is going to be 62.5 cm. Were also given the density for Mercury. Okay, the density is equal to 13.55 g per male leader. And we're right that the the level is x centimeters. We don't know what that is. Okay, So what we can do here is because we know glycerol, the level of the arm to the gas is 62.5 centimeters. We can go ahead and take that 62. centimeters And multiply it by the density of Glycerol divided by the density of Mercury. Okay, so we're going to get 1.26 g per mil leader divided by 13.55 g per millimeter. This gives us 5. centimeters which is also millimeters of mercury. So when we write that we can write 58. millimeters Of Mercury. So we want to go from mm of mercury to atmosphere. So we need to convert this. Okay, and the way that we're going to convert it is by using the conversion in one atmosphere, there is 760 of mercury. So our units cancel our mm of mercury cancel and we're left with unit of atmosphere. And when we do this calculation we get 0. atmospheres. Okay, so what do we do with this number? So we know that the pressure of the gas is going to be less than the atmospheric pressure given. So our pressure of the gas is going to equal the atmospheric pressure. So 0.942 atmospheres minus 0. Atmospheres. That we just calculated. And so the final answer is that the pressure of the gas is equal to zero 865 atmospheres. And this is going to be the final answer. That's it for this problem. I hope this was helpful