A 3.00- L tank contains air at 3.00 atm and 20.0^∘ C . The tank is sealed and cooled until the p

A 3.00- L tank contains air at 3.00 atm and 20.0^∘ C . The...

A $3.00-$ L tank contains air at 3.00 atm and $20.0^{\circ} \mathrm{C} .$ The tank is sealed and cooled until the pressure is 1.00 atm. (a) What is the temperature then in degrees Celsius? Assume that the volume of the tank is constant. (b) If the temperature is kept at the value found in part (a) and the gas is compressed, what is the volume when the pressure again becomes 3.00 atm?

Key Concepts

Ideal Gas Law

The ideal gas law (PV = nRT) is a fundamental equation that relates the pressure, volume, and temperature of a gas to its amount (in moles), with R being the gas constant. This relationship is crucial for solving problems where one or more of these variables changes under different conditions (such as heating, cooling, expansion, or compression). Even when the process involves constant volume or temperature, the ideal gas law underpins how the other variables adjust accordingly.

Constant Volume Processes

In a constant volume process, the volume does not change, which simplifies the ideal gas law to a direct proportionality between pressure and temperature (P/T = constant). When the temperature of the gas is altered while its volume remains fixed, the pressure changes in direct proportion to the change in temperature (measured in Kelvin). This concept is particularly useful in understanding how cooling or heating affects the pressure of the gas in a sealed container.

Isothermal Processes

An isothermal process is one in which the temperature of the system remains constant. For an ideal gas undergoing such a process, the product of pressure and volume remains constant (P×V = constant) as per Boyle’s Law. This means that if the pressure of the gas increases, its volume must decrease proportionally (and vice versa) so that the overall energy, specifically the temperature-dependent part, remains unchanged.

Transcript

00:04 All right, the problem is you have a tank with a volume of 3 liters and you have some gas in it such that the atmosphere, the atmospheric pressure in the gas is 3 aft, and the temperature of gas is 20 degrees celsius.

00:24 And the first part of the question asked for when you change, when you seal the tank, which means you don't allow, you don't allow the, you don't allow the, the volume be changed and you cool the temperature of the time down such that the pressure drops from three atmosphere to one atmosphere and you have to compute what the new temperature would be.

00:51 So in this case, we have the volume being constant, which means that we can start from the ideal gas law, which is pv equals to nrt, where p is the volume is the first.

01:09 Pressure v is the volume and is the number of more gas, r is a constant we call gas constant, and t is the temperature in kelvin.

01:19 We can rearrange this equation such that i have all constants push to one side and all variable pushed into the other side.

01:28 And in this case, since the tank is sealed, the volume is a constant and the number of more gas is a constant.

01:36 And so what i'm going to do is to divide both sides by v and by v and t.

01:47 So when i divide both sides by v, i get nr over v on one side.

01:53 And when i divide both sides by t, i get p over t on the other side.

01:57 So this is the equation that i get when i divide both sides by v times t, like that.

02:03 And since nr and t now are constant, then the right -hand side of the equation is a constant.

02:11 And that would mean that the left -hand side of the equation is also a constant, and the other hand side contains all the variables that i want.

02:19 So since the left -hand side is a constant, i can write p -1 over t -1 is equal to p -2 over t -2, where p -1 is the initial pressure, t -1 is the initial temperature, and et -seter.

02:32 P -2 is the final pressure, and t -2 is the final temperature.

02:35 So we arrange this equation a little bit, and you will get t1, and you will get t2, which is the final temperature, which is the quantity that we want, is equal to the initial temperature multiplied by the ratio between p2 and p1.

02:51 Because this is true.

02:53 This equation holds two...

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