An 11.5-mL sample of liquid butane (density = 0.573 g>mL) is evaporated in an otherwise empty container at a temperature of 28.5 °C. The pressure in the container following evaporation is 892 torr. What is the volume of the container?

Video transcript

Hello everyone. Today we have the following problem. An 11.5 mL sample of liquid butane with the density of 0.573 g per milliliter is evaporated in an otherwise empty container at a temperature of 28.5 °C. The pressure in the container falling evaporation is 892 tor what is the volume of the container? So recalling that under certain conditions of pressure and temperature, the ideal gas law can determine the density of the gas. And so we have the density and we can write that equation as density is equal to mass divided by volume and rearrange that to solve for the mass of the liquid. So you have the mass of the liquid is equal to density multi applied by the volume. So as given to us in the question, stem density is 0.573 g per milliliter. And our volume given to us as well as five point is 11.5 mL, units of milliliters canceled out and we are left with 6.59 g. Now, we can determine the number of moles of the liquid butane present. Am I doing that? We have our 6.59 g that will be multiplied by the molar mass of liquid butane butane, which is one more is equal to 58.12 g. Units of grams canceled out and we are left with one 0.113 moles. So before solving for our volume, we can convert the temperature of Kelvin. So we were given a temperature of 28.5 °C. So to convert that to Kelvin, we take the degrees of Celsius and we simply multiply by 2 73.15 to arrive at 301.7 degrees Kelvin. And then we can also convert the given pressure to atmospheres. We were given 892 tour which if we convert to atmospheres, we can simply use the conversion that one atmosphere as you go to 760 tor when units of Tor cancel out, we live with 1.17 atmospheres. And using the gas equation which is pressure times volume is equal to the number of moles multiplied by a gas constant, multiplied by the temperature. We can rearrange this such that our volume is equal to N RT divided by our pressure. So the number of moles that we solve for or was 0.1 13, the gas cost that can be found in reference text and this will be 0.082 liters times atmospheres divided by moles times Kelvin. And then our temperature that we saw before was 301.7 degrees Kelvin. We divide this bar pressure that we converted to atmospheres, which is 1.17 atmospheres. Um We cancel some units. We arrive at an answer or a volume of 2.8 39 leaders or answer choice be as our answer overall, I hope it helped. And until next time.