Helium gas with a volume of 2.05 L, under a pressure of 0.135 atm and at 37.0^∘ C, is warmed un

Helium gas with a volume of 2.05 L, under a pressure of...

Helium gas with a volume of $2.05 \mathrm{~L},$ under a pressure of 0.135 atm and at $37.0^{\circ} \mathrm{C}$, is warmed until both pressure and volume are doubled. (a) What is the final temperature? (b) How many grams of helium are there? The molar mass of helium is $4.00 \mathrm{~g} / \mathrm{mol}$.

Key Concepts

Combined Gas Law

The combined gas law unifies Boyle's, Charles's, and Gay-Lussac's laws, and it expresses the relationship between two states of a fixed amount of gas as (P1V1)/T1 = (P2V2)/T2. This law is essential when a gas undergoes changes in pressure, volume, and temperature simultaneously, allowing one to calculate the final state given the initial conditions.

Temperature Conversion

Temperature conversion is integral in applying the gas laws because these laws require temperatures to be in an absolute scale (Kelvin). Converting from degrees Celsius to Kelvin by adding 273.15 is a basic yet critical step to accurately determine gas properties using the ideal or combined gas laws.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry and physics that relates the pressure, volume, temperature, and number of moles of a gas through the equation PV = nRT. This equation is used to predict the behavior of an ideal gas and is crucial for solving problems involving gas properties under varying conditions.

Molar Mass

Molar mass is the mass of one mole of a substance and serves as the bridge between the number of moles and the mass of a gas. It enables the calculation of the mass from the number of moles (or vice versa) once the ideal gas law or other related calculations provide the amount of substance present.

Transcript

00:01 Hi there, so for this problem we have an alien gas with a volume that is given.

00:05 So that volume, let's label that as p1, is 2 .05 liters under an initial pressure of 0 .135 atmospheres.

00:17 And at a temperature, let's label that as temperature 1, that is 37 celsius degrees.

00:22 Now what we need to do first is to transform this temperature that is in celsius degrees to kelvin.

00:28 So to this we need to add 273, so let's add to this that.

00:40 So then this will be 37 plus 273.

00:46 So that will give us 310 kelvin.

00:57 And then with this, we can proceed to do part a of this problem, which is what is the final temperature.

01:06 So for this we can use the following equation, that is that the pressure 1 times the volume 1, this, divided by the temperature 1 is equal to the pressure 2 times the volume 2, this, and divided by the temperature 2.

01:19 Now consider that in this case we are told that the pressure doubles.

01:24 So the pressure 2 divided by the pressure 1 is equal to 2, and the volume 2 divided by the volume 1 is equal as well as 2.

01:32 So what we need to do in here is to solve for the temperature 2, so that will give us that in temperature 1, this times the pressure 2 divided by the pressure 1, this times the volume 2 divided by the volume 1.

01:45 Now we substitute the values in here, so that will be the value that we just obtained from before, 310.

01:51 Then this times, in here we will have 2 and 2, so that will give us 4 in here...

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