∙∙(a) Calculate the mass of nitrogen present in a volume of 3000 cm^3 if the temperature of the

step 1 This is chapter 15, problem number 68. We're asked to calculate the mass of a nitrogen gas, give

∙∙(a) Calculate the mass of nitrogen present in a volume of...

$\bullet$$\bullet$ (a) Calculate the mass of nitrogen present in a volume of 3000 $\mathrm{cm}^{3}$ if the temperature of the gas is $22.0^{\circ} \mathrm{C}$ and the absolute pressure is $2.00 \times 10^{-13}$ atm, a partial vacuum easily obtained in laboratories. The molar mass of nitrogen $\left(\mathrm{N}_{2}\right)$ is 28.0 $\mathrm{g} / \mathrm{mol} .$ (b) What is the density (in $\mathrm{kg} / \mathrm{m}^{3} )$ of the $\mathrm{N}_{2} ?$

$\bullet$$\bullet$ (a) Calculate the mass of nitrogen present in a volume of
3000 $\mathrm{cm}^{3}$ if the temperature of the gas is $22.0^{\circ} \mathrm{C}$ and the
absolute pressure is $2.00 \times 10^{-13}$ atm, a partial vacuum easily
obtained in laboratories. The molar mass of nitrogen $\left(\mathrm{N}_{2}\right)$ is
28.0 $\mathrm{g} / \mathrm{mol} .$ (b) What is the density (in $\mathrm{kg} / \mathrm{m}^{3} )$ of the $\mathrm{N}_{2} ?$

Key Concepts

Molar Mass and Mass Calculation

Molar mass is the mass of one mole of a substance and is essential for converting between the number of moles and the mass of a given gas. It allows for the determination of the actual mass from the calculated moles obtained using the Ideal Gas Law.

Gas Density

Density is defined as mass per unit volume and is a useful property in characterizing gases. Calculating the density involves dividing the mass of the gas by its volume, and unit conversions are often required to ensure both mass and volume are expressed in consistent units, such as kilograms and cubic meters.

Ideal Gas Law

The Ideal Gas Law (PV = nRT) is fundamental in relating the pressure, volume, and temperature of an ideal gas with the amount of substance in moles. It is widely used in thermodynamics to calculate unknown parameters of gases when the other variables are known, making it integral to understanding and solving gas-related problems.

Temperature Conversion to Kelvin

For accurate gas law calculations, temperature must be expressed in Kelvin rather than Celsius. This conversion is achieved by adding 273.15 to the Celsius temperature, ensuring that the units are consistent and that the calculations based on absolute temperature are valid.

Transcript

00:01 This is chapter 15, problem number 68.

00:03 We're asked to calculate the mass of a nitrogen gas, given the volume, temperature, pressure, and the molar mass of this gas in part 8.

00:14 So if we want to calculate the mass of a nitrogen, given the volume is 3 ,000 centimeter cubed, the temperature is 22 celsius, pressure is 2 times 10 .3 .3 cm.

00:30 Of negative 13 atmosphere, and the molar mass of the nitrogen gas is given as 28 grams per per mole.

00:40 Now, from the ideal gas law then, to answer to part a, pv equals nrt, right? first, we need to figure out how many moles of this gas there are in this volume so that afterwards we can calculate the total mass, right? so first thing, first, we need to figure out what n is.

00:59 In order to do that if we divide both sides by r times t then the number of moles equals p b over r t now we need to do a lot of unit conversions here the pressure first of all let's write the pressure in terms of pascals right so two times center for negative 13 atmosphere is what's given to us as we as you know um 1 .013 times 10 to the power 5 pascels per atmosphere, right? we can do the conversion.

01:36 And then we're also given 3 ,000 centimeter cubed in order to, well, convert that to meter cubed.

01:42 We're going to multiply 3 ,000 by 10 to the power negative 6 so that it would give us the meter cube, right? divided by r would be 8 .315 joules per mole, calvin, and the gas constant, right? and we need to convert our temperature, which is given to us in celsius, to kelvin.

02:10 In order to do that, we're just going to add 273 to it, right? so we can write this in terms of kelvin.

02:19 So then the number of moles we have is 2 .48 times 10 to 0 .14 for this gas.

02:28 Since we know how many moles of this gas we have, now we can calculate the mass, right? so the mass is going to be equal to number of moles times the molar mass of this gas...

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