The pressure of a sample of CH₄ gas (6.022 g) in a ...

Question

The pressure of a sample of CH₄ gas (6.022 g) in a 30.0 L vessel at 402 K is ________ atm.

Accepted Answer

Hello. Everyone in this video, we're going to be working with a gas solved problem. So whenever we deal with gas laws, we want to keep in mind the ideal gas law because that is going to be the ones that's easiest and most applicable for all the gas law equations. So, first we want to separate our given information. So when reading our problem, we can see that we have 20 liters And that's going to have 41.3 g of chlorine gas. So she here in the gaseous state, then we have a temperature of 3 15 Calvin's. So let's see, let's see if this fits into and we can use the PV equals NRT is our deal guests equation. So PV equals N. R. T. Good. And put rectangle on that just because that's what we're aiming for. So, from the information we can see that we have a volume. We have what we could make into a mole. So that's the end and we have a T. Value. So, let's see here. So, we have we do not have a P value but we do have a V. We have an end. We have the R. And we have the T. So yes, we can use this. This PV equals NRT equation. So, we're trying to find here in this problem will be our P which is pressure. All right. So Our our value is going to be our constant. This capital R. That's equal to 0.08206. The units of being ATM times later all over kelvin's times mold. So whenever we're putting values into this PV equals Nrt equation. We want all the information given to have units that match up to this R value here. So we have a T. M. Leaders, kelvin and moles. So we have our leaders. That's great. Let's see here. This we don't have because we want an end to be our moles and we have in grams but we can do some simple dimensional analysis and get that done. Then we have a temperature at kelvin. This is kelvin. So that's great. So Doing that dimensional analysis. So we have let's see our 41. g of chlorine gas. So we want to do dimension analysis to get that into molds. So on the side here 1st or on top, We want to recognize that chlorine gas is cl two. So we have to Unit are two atoms of chlorine and each individual atom of chlorine that weighs in At 35. g. So adding that into my calculator, I will get the value of city . g per moles. That would be my converted factor to plug in to go from grams into moles. So the denominator I put 70.9 grams through everyone more. Alright, you see that the grounds will cancel out nicely leaving us with the moles of chlorine gas. So putting that into back out clear. I get 0.5 8 to Moles of Cl two. All right. No, I want to also manipulate this equation because we're solving only for the P. I want to go identified each side by RV. So go ahead and do that divide each side by V. You see that these were cans out leaving us with equation which is P. So the pressure you're going to N R. T over V. Let's plug in our values. So our mold we just calculated for is 0.58-5 moles. Then we're going to notify that our guests are constants, the capital art. 0.8206 units. Speak A. T. M. Times leader. All over Calvin times mold. And then we're also able to find that bar temperature And we have we are given that of 3 15 Kelvin's. That's all going to be divided by the volume. Our volume is 20 L. You can see now, but the moles cancel helped Calvin's cancel out and leaders cancel out leaving us with just the pressure unit of a T. M. So bring that into my calculator. I'll get the value of 0. A. T. M. And that's going to be my final answer for this problem. Thank you for watching

Difficoltà con General Chemistry?

The pressure of a sample of CH₄ gas (6.022 g) in a 30.0 L vessel at 402 K is ________ atm.

Guarda il prossimo

Set di pratica su The Ideal Gas Law

Difficoltà con General Chemistry?

The pressure of a sample of CH4 gas (6.022 g) in a 30.0 L vessel at 402 K is ________ atm.

Guarda il prossimo

Set di pratica su The Ideal Gas Law

The pressure of a sample of CH₄ gas (6.022 g) in a 30.0 L vessel at 402 K is ________ atm.