### Video Transcript

Airtight bags containing food that are taken to high elevations become puffed up because the air inside the bags has expanded. A bag of pretzels was packed at a pressure of 1.00 atmospheres and at a temperature of 25.0 degrees Celsius. The bag is taken to a summer picnic in Santa Fe, New Mexico, where the air temperature is 35.0 degrees Celsius. The volume of the air in the bag is 1.24 times its volume when packed. What is the pressure of the air at the picnic?

Weβll label the pressure of the air at the picnic π sub π. And to solve for it, weβll assume that the air inside the bag of pretzels is an ideal gas. In this scenario, at the outset, we have a bag of pretzels being packed at a certain pressure, temperature, and volume. Weβll label those initial values π sub π, π sub π, and π sub π. π sub π weβre told is equal to 1.00 atmospheres of pressure and π sub π the initial temperature is 25.0 degrees Celsius.

After this bag is packed, itβs transferred to a new location. Under these new atmospheric conditions, the bag expands. So it looks all puffed up. Weβre told that after the bag expands, its final value to its initial volume equals 1.24. And weβre told that the final temperature of the air in the bag is 35.0 degrees Celsius.

Knowing all this, we want to solve for the final pressure that the bag experiences. The ideal gas law can help us solve for this final pressure. This law says that if we take the pressure times the volume of an ideal gas, then that product is equal to the number of moles of the gas times the gas constant called π
multiplied by the temperature of the gas π in kelvin.

We can use this relationship to express pressure volume and temperature both for the initial as well as the final condition of our bag. When we write out those two equations, we notice that the number of moles of gas in the bag as well as the gas constant are the same in both these two equations.

Rearranging, we can say that π times π
, this constant, is equal to ππ ππ over ππ, which itself is equal to ππ ππ over ππ. Itβs this second written down equality that weβll focus on to solve for π sub π, the final pressure of the air. Rearranging once more, we find that the final pressure is equal to the initial pressure times the ratio of the final to the initial temperature multiplied by the ratio of the initial to the final volume of the air in the bag.

Since the problem statement tells us the initial pressure as well as the initial and final temperatures and the ratio of volumes of the bag, weβre ready to plug in and solve for π sub π. As we do, one unit conversion weβll want to make is to convert our two temperatures from degrees Celsius to the kelvin scale.

To solve for that value, weβll take the temperature in degrees Celsius and add 273.15 to it. Plugging all these values in as well as entering one over 1.24 for π sub π over π sub π and 1.00 atmospheres for the initial pressure, when we calculate π sub π, we find to three significant figures 0.834 atmospheres. Thatβs the final pressure the bag must experience in order to expand this much under this temperature difference.