Five grams of nitrogen gas at an initial pressure of 3.0 atm and at 20°C undergo an isobaric expansion until the volume has tripled.

b. What is the gas temperature after the expansion (in °C)? The gas pressure is then decreased at constant volume until the original temperature is reached.

Question

Five grams of nitrogen gas at an initial pressure of 3.0 atm and at 20°C undergo an isobaric expansion until the volume has tripled.

b. What is the gas temperature after the expansion (in °C)? The gas pressure is then decreased at constant volume until the original temperature is reached.

Accepted Answer

Hey, everyone. Let's go through this practice problem. Consider a container filled with 15 g of methane gas at a pressure of 2.2 atmospheres and a temperature of 30 degrees Celsius. The gas undergoes an isobaric expansion during which its volume increases by a factor of four, determine the final temperature of the gas in degrees Celsius. And we have four options to choose from option, a 1320 degrees Celsius, option B 1212 degrees Celsius, option C 303 degrees Celsius and option D 939 degrees Celsius. Ok. So the problem is asking us for a final temperature, which means we're going to want to find a relationship between the initial and final variables before and after the volume increase for this gas. And one way we can compare the variables before and after the expansion is by using the ideal gas law, which states that the initial pressure, which I'm going to, I'm gonna use the subscript one to refer to the initial variables, but the initial pressure multiplied by the initial volume divided by the initial temperature is equal to the final pressure. And denoted by a two subscript multiplied by the final volume divided by the final temperature. And the problem is asking to, for us to determine the final temperature. So I'm going to solve this equation for the final temperature T sub two, we can find a formula for T sub two quite simply algebraically by multiplying both sides of the equation by T sub two to get T sub two in the numerator on its own on the left hand side of the equation, and then we can get T sub two all by itself by multiplying both sides of the equation by T sub one. And then dividing both sides of the equation by P sub one V sub one. So what we find is that T sub two is equal to P sub two V sub two multiplied by T sub one all divided by P sub one multiplied by V sub one. Now, there are a few different ways we can simplify this first. The problem tells us that the expansion is isobaric which means that the pressure is constant and doesn't change. So P sub one is going to be equal to P sub two. So the pressures can then cancel out. And our formula for T sub two can be more simply written as T sub two equals V sub two, T sub one divided by V sub one. The problem also tells us that the volume increases by a factor of four. In other words, V sub two. Is equal to four multiplied by V sub one. Let's plug this in to V sub two. We'll substitute that for V sub two in our temperature equation. So instead that's four multiplied by V sub one, multiplied by T sub one divided by V sub one. So now we have a V sub one in both the numerator and the denominator, so they can cancel out. So all we're left with is on the right hand side of the equation is four multiplied by T sub one. So in other words, the final temperature is equal to four times the initial temperature. So let's do some math here, T sub two is equal to four multiplied by T sub one. Now, the initial temperature is given to us in the problem as 30 degrees Celsius. But it's worth noting that in order for the combined ideal gas law to work, we need to have our temperature in units of Kelvins rather than degrees Celsius. 30 sub one is equal to 30 plus 273 Kelvins. And that's how we convert from degrees Celsius into Kelvins. And so our initial temperature is 303 Kelvins. So T sub two is equal to four multiplied by Kelvins. And if we put this into a calculator, then we find the final temperature has a value of 1212 Kelvins. However, if we look at our multiple choice options, we can see that all four of the options given to us are in units of degrees Celsius. So we're going to have to take the answer we just found and then subtract 273 again to get it back into degrees Celsius that 1212 minus 273 degrees Celsius, which gives us an answer of about 939 degrees Celsius. So that means that this is our final temperature, which means that this is our answer to the problem. If we look at our multiple choice options, we can see that this agrees with option B which states a temperature of 939 degrees Celsius. So option D is our answer to the problem. And that's all for this video. I hope this video helped you out. If you'd like more practice, please check out some of our other tutoring videos which will give you more experience with these types of problems, but that's all for now. And I hope you all have a lovely day. Bye bye.

Five grams of nitrogen gas at an initial pressure of 3.0 atm and at 20°C undergo an isobaric expansion until the volume has tripled. b. What is the gas temperature after the expansion (in °C)? The gas pressure is then decreased at constant volume until the original temperature is reached.

Verified Solution

Key Concepts

Ideal Gas Law

Isobaric Process

Charles's Law