A sealed container holds 3.2 g of oxygen at 1 atm pressure and 20°C. The gas first undergoes an isobaric process that doubles the absolute temperature, then an isothermal process that halves the pressure. What is the final volume of the gas in L?

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Hey, everyone. Let's go through this thermodynamics problem. In a thermodynamic experiment, five g of nitrogen gas at 400. Kelvins is placed inside a piston cylinder of volume two multiplied by 10 to the power of negative five cubic meters under a pressure of two multiplied by 10 to the power of five pascals. First, the temperature of the gas is increased by 40% while its volume remains constant. Then the pressure is reduced by 30% at a constant temperature. Calculate the final volume of the nitrogen gas. We are given four, multiple choice options. Option A 1.4 multiplied by 10 to the power of negative five cubic meters. Option B 2.9 multiplied by 10 to the power of negative five cubic meters. Option C 4.1 multiplied by 10 to the power of negative five cubic meters. And option D 6.7 multiplied by 10 to the power of negative five cubic meters. So to recap, we're looking for a final volume of the gas based on other variables like the temperature, the mass and its pressure. So this is going to be a case where we'll want to use the ideal gas law, more specifically, we can use the combined ideal gas law, which states that the pressure multiplied by the volume divided by the temperature of a gas is constant. Even through transformations and processes that change any of these variables. We can use this equation to compare the variables between two different states of the gas. And based on the way the problem is laid out, we essentially have three different processes to work from or three different states of the gas. I'm gonna call state A, the original state of the gas where the pressure P sub A is given to us as 2. multiplied by 10 to the power of five pascals. The volume V sub A is given to us as 2.0 multiplied by 10 to the power of a negative five cubic meters. And the temperature T sub A is given to us as 400 Kelvins. These are the initial variables for the gas. So I'm going to say that these three variables correspond to state A state B will refer to the variables after the temperature of the gas is increased by 40%. We're not told how the pressure changes in state B. They'll say that P A B, the pressure of state B is unknown, but we are told that the volume remains constant. So V sub B is equal to V sub A here. But we're told that the temperature of the gas is increased by 40%. So T sub B is equal to T sub A plus 40% of bay. So plus 0.4 multiplied by T sub A and we can put this into a calculator to find what that temperature is. So 400 Kelvins plus 0. multiplied by 400 Kelvins. And if we put that into our calculator, then we find a new temperature for T sub B as Kelvins. Lastly, the third state state C is when the pressure is reduced by 30% at a constant temperature. So the new pressure P sub C is equal to P sub B minus 30% of what P sub B originally was. So minus 300.3 P A B V sub C, the final volume is unknown and is in fact what the, what the problem is asking us to find anyway. And we're told that the temperature is kept constant during this process. So T sub C is equal to T sub B. Let's first use the combined ideal gas law to solve for P sub B. And then once we have all the variables for the B state, then we can use the combined ideal gas law again for states B and C to then solve for the final volume of the gas. So let's start by comparing the first two states. So P sub A multiplied by V sub A divided by T sub A is all equal to P sub B multiplied by V sub B all divided by T sub B. And the problem tells us that for the first process, the volume is kept constant as we discussed earlier. The sub A is then equal to V sub B. So those variables cancel out and we're just left with P sub A divided by T sub A is equal to P sub B divided by T sub B. So let's then solve this equation for P sub B by multiplying both sides of the equation by T sub B. So P sub B is equal to P sub A multiplied by T sub B divided by Ti Sabe. So now let's plug in the values we have P sub B is equal to the pressure at A which as we discussed earlier is given in the problem as 2. multiplied by 10 to the power of five pascals and then multiplied by the temperature T sub B. She was 560 calvins divided by T sub A 400 Kelvins. And if we put this into our calculator, then we find that the pressure at state B is 2. multiplied by 10 to the power of five pascals. So now we know what peace is. So now that we know all of the variables for state B, now we can use the combined ideal gas law again to compare the variables between state B and C. So P sub B multiplied by the sub B divided by T sub B is equal to P sub C multiplied by V sub C divided by T sub C. And as the problem tells us for the second process, the temperature is kept constant. So T sub B is equal to T sub C and those cancel out. So we're left with P sub B multiplied by V sub B is equal to P sub C multiplied by V sub C. We're trying to solve for the vinyl volume here. So let's solve this for V sub C by taking both sides of the equation and dividing it by P sub C. So V sub C is equal to V sub B multiplied by P sub B divided by P sub C. Finally let's just plug everything into a calculator. So V so V sub B is equal to or Vub C is equal to V sub B which we know is the same as V set. A 2.2 multiplied by 10 of the power of negative five cubic meters two multiplied by 10. The power of negative five cubic meters multiply applied by the pressure of point B. So 2.8 multiplied by 10 of the power of five past scales divided by the pressure at C which we talked earlier is P sub B minus 0.3 P sub B. So in the denominator here is 2. multiplied by 10 of the power of five pascals minus 0.3, multiplied by 2.8, multiplied by 10 to the power of five pascals. I'm gonna make this dot A little more pronounced. And now if we were to plug this into a calculator, we'll find a volume of C of about 2.9 multiplied by 10 to the power of a negative five cubic meters. And so that means that this is our answer to this problem because this is the final volume of the gaps. And if we look at our options, we can see that option B is this app answer 2. multiplied by 10 to the power of negative five cubic meters. And that is our answer to this problem. Thank you for watching. I hope this video helped you out. If you'd like some more experience with these types of problems, please check out some of our other videos which should give you more practice and that's all for now. I hope you all have a lovely day. Bye bye.

A sealed container holds 3.2 g of oxygen at 1 atm pressure and 20°C. The gas first undergoes an isobaric process that doubles the absolute temperature, then an isothermal process that halves the pressure. What is the final volume of the gas in L?

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Key Concepts

Ideal Gas Law

Isobaric Process

Isothermal Process