In this video blog post, we'll do several __gas law practice problems__. The first two examples only contain "1 situation."

These are __Ideal Gas Law__ conditions -- only 1 situation.

**Ideal Gas Law Examples**

*ex: * A sample of H

_{2}gas occupies a volume of 8.56 L at a temperature of 0°C and a pressure of 1.5 atm. How many moles of hydrogen are present?

_________

**answer:**➞ **PV = nRT** ➞ n = PV / RT

➞ n = (1.5 atm) (8.56 L) **/** (0.0821 L^{.}atm/K^{.}mol) (273 K)

➞ **n = 0.57 mol H _{2}**

Here's the same solution, but with a natural flow...

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*ex:* Calculate the volume of 0.845 mol of nitrogen gas at a pressure of 699.2 torr and a temperature of 315 K.

_________

**answer:**➞ we have to convert pressure from torr to atm

➞ given:

n = 0.845 mol N_{2}

T = 315 K

V = ?

P = 699.2 torr

(699.2 torr / 1) ^{.} (1 atm / 760 torr) = 0.9200 atm

➞ **PV = nRT ** ➞ V = nRT / P

➞ V = nRT / P = (0.845)(0.0821)(315) **/** 0.9200 = **23.8 L N _{2}**

And, here's an easier-to-follow view of the above process:

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* TIP*:

**PV = nRT** problems don't typically have a "situation 1" and "situation 2" occurring. There's usually just one situation.

"Before / After" type problems usually require Boyle's Law, Charles' Law, Avogadro's Law, Gay-Lussac's Law, or the Combined Gas Law.

Let's try some!...

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**Charles' Law Example**

*ex:* A sample of methane gas with a volume of 38 mL at 5°C is heated to 86°C at constant pressure. Calculate its new volume.

_________

**answer:**➞ **V _{1}** = 38 mL ,

**T**= 5 + 273 = 278 K

_{1}➞ **V _{2}** = ?? mL ,

**T**= 86 + 273 = 359 K

_{2}... How come it wasn't necessary to convert V_{1} and V_{2} into liters (L) ?

Because * "R"* (in PV = nRT) is equal to 0.0821 L

^{.}atm / K

^{.}mol, but

*is not being used in*

**"R"**__Charles' Law__!!

**:-)**

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**Combined Gas Law Example**

** ex:** Diborane gas (B

_{2}H

_{6}) has a pressure of 345 torr at a temperature of -15°C and a volume of 3.48 L. If conditions change such that the temperature is 36°C and the pressure is 468 torr, what will be the volume of the sample?

_________

**answer:**__situation 1__:

P_{1} = 345 torr

T_{1} = -15 + 273 = 258 K

V_{1} = 3.48 L

__situation 2__:

P_{2} = 468 torr

T_{2} = 36 + 273 = 309 K

V_{2} = ?? L

➞ we'll use the * Combined Gas Law*,

P

P

_{1}V_{1}/ T_{1}= P_{2}V_{2}/ T_{2}** V _{2} = 3.07 L B_{2}H_{6}**

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**Ideal Gas Law Practice Problem**

*ex:* A gaseous sample contains 0.35 moles of Argon at a temperature of 13°C and a pressure of 568 torr. If it's heated to 56°C and a pressure of 897 torr, calculate the change in volume that occurs.

_________

**answer:**➞ this one's tricky because at first we see two pressures...

However, n_{1} = n_{2} = remains unchanged and we're asked to find the change in volume (ΔV).

To find ΔV = V_{2} - V_{1}, **we'll need to use PV = nRT twice...**

We will need to convert our pressures from torr to atmospheres (atm), and then set-up two separate PV = nRT equations:

**➞ P _{1}V_{1} = nRT_{1}** and

**P**

_{2}V_{2}= nRT_{2}Here's the work that's involved...

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That's it for entry #3 from **SECTION 5 - Gases**.

In my next video from this Section ("Chapter"), we'll discuss Gas Stoichiometry and how to derive Molar Mass and Density from the Ideal Gas Law.