Question

PLEASE HELP!! Answer whatever ones you can please and thank you!!!!

1. A gas with a volume of 4.0 L at a pressure of 2.03 atm is allowed to expand to a volume of 12.0 L. What is the pressure in atm in the container if the temperature remains constant. Please show work (including the equation you used).

2.The volume of a gas is 200 mL at 350.0 kPa pressure. What will the volume be when the pressure is reduced to 120.0 kPa, assuming the temperature remains constant. Please show all work for full credit. Please show work (including the equation you used).

3.What is the original pressure of a bike tire if the current pressure is 150 atm and the volume changed from 1000 mL to 700 mL. Please show work (including the equation you used).

Answer 1

These problems can be solved using the gas laws. The combined gas law states that PV/T = k, where P is the pressure, V is the volume, T is the temperature (in Kelvin), and k is a constant. From this relationship, you can figure out the change in pressure, volume, or temperature of a gas provided that one of those three parameters remains unchanged. This will be shown in the problems given.

1. Here, the pressure and volume of a gas change while the temperature remains constant. Since PV/T = k, if T is also constant, then the relationship becomes PV = k. In other words, the pressure and volume of a gas at a constant temperature are inversely proportional to each other. This can be expressed mathematically as
`P_1V_1=P_2V_2` (also called Boyle's Law), where P₁ and V₁ denote the initial pressure and volume, and P₂ and V₂ denote the final (or changed) pressure and volume, respectively.

We are given a gas with an initial pressure, P₁, of 2.03 atm and an initial volume, V₁, of 4.0 L. We are also told that the gas expands to a final volume, V₂, of 12.0 L. So, we are looking for P₂: The pressure of the gas when the volume has expanded from 4.0 to 12.0 L. From Boyle's Law,

`P_2=(P_1V_1)/(V_2)`.

Thus,

`\[P_2=\frac{\left ( 2.03 \text{ atm} \right )\left ( 4.0 \text{ L} \right )}{\left ( 12.0 \text{ L} \right )}\] = 0.67 atm.`

Since the pressure and volume of a gas are inversely proportional, as one increases, the other decreases, and vice versa.

2. This is another Boyle's Law problem, but this time, we have to figure out what the final volume will be when the pressure of a gas is changed. Since the pressure is reduced, we can deduce that the volume must increase (notice the parallel with the previous question). We again turn to Boyle's Law, and this time we solve for V₂:

`V_2=(P_1V_1)/(P_2)`

from which we obtain

`\[V_2=\frac{\left ( 350.0 \text{ kPa} \right )\left ( 200 \text{ mL} \right )}{\left ( 120.0 \text{ kPa} \right )}\] = 583 mL.`

It's not clear whether 200 mL (V₁) should be assigned three significant figures or one. I gave the answer to three significant figures; the answer to one significant figure would be 600 mL.

3. We use Boyle's Law once more here, but our word problem has us working backwards (it's also not explicitly mentioned that temperature remains constant, but it should be reasonable to assume so). The current pressure of the gas inside the bike tire is 150 atm at a volume of 700 mL . But we want to know what the initial pressure of the gas was at a volume of 1000 mL. In other words, we want to solve for the P₁ in Boyle's Law:

`P_1=(P_2V_2)/(V_1).`

Since the volume of the gas had decreased from 1000 mL (V₁) to 700 mL (V₂), and the pressure and volume of a gas at a constant temperature are inversely proportional, we would expect the pressure of the gas to have increased; that is, our P₁ should be less than 150 atm.

`\[P_1=\frac{\left ( 150 \text{ atm} \right )\left ( 700 \text{ mL} \right )}{\left ( 1000 \text{ mL} \right )}\] = 105 atm.`