A sample of dichloromethane gas (CH 2 Cl 2 ) occupies 32.6 L at 310 K and 5.30 atm. Determine its volume at STP?

Here in this practice question it says, a sample of dichloromethane gas, CH2CL 2 occupies 32.6 liters at 310 Kelvin and 5.30 atmospheres. Determine its volume at s t p. Alright. So in this question, they're giving us a volume, a temperature, and a pressure. But now they're asking us for a second and new volume, so v 2, which would mean that this is v 1. Then they're giving us STP conditions now, so this would give us t 2 and p 2, which would mean that this is v 1, t 1 and p 1. Looking at the ideal gas law, we know we have p v equals nrt. Now in our videos entitled ideal gases derivations, I told you that if you have a variable with 2 different values, then then we're gonna focus on those variables because they're changing. So we have 2 pressures, 2 volumes, 2 temperatures. Moles are not being discussed because they're being held constant, r is our constant. Bring over the t to the other side and this would give us p 1voneovertoneequalsp2v2overt2. So remember this is called our combined gas law, and the way we got it is through the videos where we talk about ideal gas law derivations. So if you haven't taken a look at those videos, I highly suggest you go back and take a look. That's how we're able to derive this combined gas law formula from the ideal gas law. All we do now is we plug in the information we have and solve for the one missing variable, which is v 2. So p 1, pressure 1 is 5.3 atmospheres, then we have v 1 which is 32.6 liters over t 1 which is 3 10 kelvin equals so we have 1 atmosphere for p 2 because it's standard temperature and pressure. V 2 is what we're looking for and then temperature 2 at s t p is 273.15 Kelvin. So at this point now we're trying to cross multiply these 3 and we're gonna cross multiply these 3. Bringing that down, that's gonna give us v 2 times 1 atmosphere times 310 Kelvin equals 5.30 atmospheres times 32.6 liters times 273 0.15 Kelvin. Divide both sides here by 1 atmosphere and 3 10 Kelvin. When we do that we'll be able to isolate our v 2, cancel out and then cancel out units. Atmospheres cancel out, kelvins cancel out. So now we'll have our v 2, which will be in, liters. So we're gonna get 152 liters of our dichloromethane gas. Here our answer at the end has 3 significant figures because all of these values given to us oh, actually, 2 of these values given to us have 3 sig figs. This 310 here only has 2 sig figs. So technically, 1 50 liters would be the more accurate answer. So that would be our exact answer that we need for this

Table of contents

Skip topic navigation

1. Matter and Measurements4h 31m
2. Atoms and the Periodic Table5h 23m
3. Ionic Compounds2h 18m
4. Molecular Compounds2h 18m
5. Classification & Balancing of Chemical Reactions3h 17m
6. Chemical Reactions & Quantities2h 27m
7. Energy, Rate and Equilibrium3h 46m
8. Gases, Liquids and Solids3h 25m
9. Solutions4h 10m
10. Acids and Bases3h 29m
11. Nuclear Chemistry56m
BONUS: Lab Techniques and Procedures1h 38m
BONUS: Mathematical Operations and Functions47m
12. Introduction to Organic Chemistry1h 34m
13. Alkenes, Alkynes, and Aromatic Compounds2h 12m
14. Compounds with Oxygen or Sulfur1h 6m
15. Aldehydes and Ketones1h 1m
16. Carboxylic Acids and Their Derivatives1h 11m
17. Amines39m
18. Amino Acids and Proteins1h 51m
19. Enzymes1h 37m
20. Carbohydrates1h 41m
21. The Generation of Biochemical Energy2h 8m
22. Carbohydrate Metabolism2h 22m
23. Lipids2h 26m
24. Lipid Metabolism1h 45m
25. Protein and Amino Acid Metabolism1h 37m
26. Nucleic Acids and Protein Synthesis2h 54m

8. Gases, Liquids and Solids

Standard Temperature and Pressure

GOB Chemistry

8. Gases, Liquids and Solids

Standard Temperature and Pressure

Previous problemNext problem

Struggling with GOB Chemistry?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

A sample of dichloromethane gas (CH₂Cl₂) occupies 32.6 L at 310 K and 5.30 atm. Determine its volume at STP?

145 L

150 L

320 L

350 L

390 L

Previous problemNext problem

Verified step by step guidance

Identify the initial conditions of the gas: initial volume (V1) = 32.6 L, initial temperature (T1) = 310 K, and initial pressure (P1) = 5.30 atm.

Recall the conditions for STP (Standard Temperature and Pressure): STP is defined as a temperature of 273.15 K and a pressure of 1 atm.

Use the Combined Gas Law to relate the initial and final states of the gas: \( \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} \), where P2 and T2 are the pressure and temperature at STP, and V2 is the volume we want to find.

Substitute the known values into the Combined Gas Law equation: \( \frac{5.30 \times 32.6}{310} = \frac{1 \times V_2}{273.15} \).

Solve for V2 by rearranging the equation: \( V_2 = \frac{5.30 \times 32.6 \times 273.15}{310} \). This will give you the volume of the gas at STP.

0:33m

Watch next

Master Standard Temperature and Pressure with a bite sized video explanation from Jules

Start learning

Standard Temperature and Pressure practice set

Problem sets built by lead tutors

Expert video explanations

Go to Practice

Download the Mobile app

Privacy

Do not sell my personal information

Accessibility

Patent notice

Sitemap

Struggling with GOB Chemistry?

A sample of dichloromethane gas (CH2Cl2) occupies 32.6 L at 310 K and 5.30 atm. Determine its volume at STP?

Watch next

Standard Temperature and Pressure practice set

A sample of dichloromethane gas (CH₂Cl₂) occupies 32.6 L at 310 K and 5.30 atm. Determine its volume at STP?