What volume will 2.04 g of helium gas occupy at 100 .^∘ C and 785 mm Hg pressure?

Name: Problem 53, Chapter 13: Introductory Chemistry
Uploaded: 2020-06-04T19:17:02-08:00
Duration: 2 min 55 s
Channel: Bcrypt_Sha256$$2B$12$G.2Fsvlf/D/Uztpomcl6Kehenyybhk3.Xacinplxct80U//Oh6Kzq Bcrypt_Sha256$$2B$12$G.2Fsvlf/D/Uztpomcl6Kejwqx2Ml/Bpksr62Kuhpbllelwkwnzlk
Description: What volume will 2.04 g of helium gas occupy at 100 .^∘ C and 785 mm Hg pressure?

step 1 All right, so today we're talking about using the ideal gas loss. So our PV equals NRT in order

What volume will 2.04 g of helium gas occupy at 100 .^∘ C...

Bcrypt_Sha256$$2B$12$G.2Fsvlf/D/Uztpomcl6Kehenyybhk3.Xacinplxct80U//Oh6Kzq Bcrypt_Sha256$$2B$12$G.2Fsvlf/D/Uztpomcl6Kejwqx2Ml/Bpksr62Kuhpbllelwkwnzlk and 76 other subject Chemistry 101 educators are ready to help you.

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Key Concepts

Ideal Gas Law

The ideal gas law, represented as PV = nRT, connects the pressure, volume, number of moles, and temperature of an ideal gas. It is a fundamental principle used to solve for any one of these properties, provided the others are known and that the gas behaves ideally under the given conditions.

Molar Mass

Molar mass is the mass of one mole of a substance and is a critical concept for converting the given mass of a gas into the number of moles. This conversion is an essential first step before applying the ideal gas law in calculations involving gases.

Temperature Conversion

Temperature conversion, specifically converting Celsius to Kelvin, is crucial in gas calculations because the ideal gas law requires temperature to be expressed on an absolute scale. Using Kelvin ensures the correctness of thermodynamic equations.

Pressure Conversion

Pressure conversion involves translating pressure values from one set of units to another, such as from mm Hg to atmospheres. This standardization is necessary to ensure consistency in units when applying the ideal gas law, which depends on using a gas constant that aligns with the chosen units.

Transcript

00:01 All right, so today we're talking about using the ideal gas loss.

00:04 So our pv equals nrt in order to find missing quantities.

00:08 And so in this problem, we're told that we have, so we need to find our variables first.

00:13 So we're told that we have 2 .04 grams of helium.

00:19 And so we know that we can get from grams per moles just with some dimensional analysis.

00:24 And so we'll go ahead and convert that over.

00:27 So we'll multiply it by our conversion factor.

00:29 And so we find that in one mole we have four grams of helium.

00:35 And so we find that from the periodic table.

00:38 And then that gives us 0 .510 moles.

00:44 And so the next in our problem, we find that we have 100 degrees celsius for our temperature.

00:49 So we need to be in kelvin.

00:50 So we just add our 273 to convert between celsius and kelvin.

00:55 And we get 373 kelvin.

00:58 And then finally, we're told that our pressure, is 785 millimeters of mercury.

01:07 And again, we need that in atmospheres, so we're going to have to convert.

01:12 So we'll go ahead and start by converting our pressure here.

01:16 Oh, let me get that.

01:20 And so we know that there are 760 millimeters of mercury in one atmosphere...

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