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The vapor pressure of water at 40.0ºC is $7.34 \times 10^{3} \mathrm{N} / \mathrm{m}^{2} .$ Using the ideal gas law, calculate the density of water vapor in $\mathrm{g} / \mathrm{m}^{3}$ that creates a partial pressure equal to this vapor pressure. The result should be the same as the saturation vapor density at that temperature $\left(51.1 \mathrm{g} / \mathrm{m}^{3}\right)$

$\rho=50.8 \frac{\mathrm{g}}{\mathrm{mol}}$

Physics 101 Mechanics

Chapter 13

Temperature, Kinetic Theory, and the Gas Laws

Temperature and Heat

University of Michigan - Ann Arbor

Hope College

University of Sheffield

McMaster University

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So here we're going to use the ideal gas law. The pressure terms of volume equals the number of most times the ideal gas, constant times a temperature. Now we can say that the number of moles divided by the volume would be equaling the pressure divided by the ideal gas. Constant times t the temperature. Ah, we're going to then solve and say that this is gonna be equaling 7.34 times 10 to the third Newtons per square meter, divided by 8.31 Jules Perk, Calvin Permal multiplied by the temperature of 313.15 Calvin. And so we have that the moles per unit volume would be equaling 2.82 moles per cubic meter. And so we know that the density is equaling the number of moles per cubic per unit volume or per cubic meter times Mm ah, the Moler mass of water. And so the density would be equaling 2.82 moles. Ah, per unit volume in this case, cubic meters multiplied by the molar mass of water, 18.0 grams Permal. And this is equal in approximately 50.8 grams per cubic meter. Um and this is approximately equal to what is given 51.1 grams per cubic meter. This would be the saturation vapor density, but this would be our final answer. So it is about 0.3 grams per cubic meter off. That is the end of the solution. Thank you for one.

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