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, 150 liters would be the more accurate answer. So that would be our exact answer that we need for this

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11 Gases

Standard Temperature and Pressure

Introduction to Chemistry

11 Gases

Standard Temperature and Pressure

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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

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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: \( \frac{P_1 \cdot V_1}{T_1} = \frac{P_2 \cdot V_2}{T_2} \), where P2 = 1 atm, T2 = 273.15 K, and V2 is the volume at STP we need to find.

Rearrange the combined gas law to solve for the final volume (V2): \( V_2 = \frac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot P_2} \).

Substitute the known values into the equation: \( V_2 = \frac{5.30 \cdot 32.6 \cdot 273.15}{310 \cdot 1} \) and simplify to find the volume at STP.

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Struggling with Introduction to 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?