Question 2
The volume of a gas-filled balloon is 20.0 L at 60 atm pressure. What volume in liters will the balloon have at 30 atm?

Question 3
8.00 L of gas at standard temperature and pressure (STP) is compressed to 3 L. What is the new pressure of the gas in atm?

Question 4
If a tennis ball has a pressure of 200 atm at a temperature of 27oC, what pressure in atm will the tennis ball have if the temperature of the gas increased to 77oC?


Question 5
Exactly 5.00 L of air at -23oC is warmed to 27o What is the new volume in liters if the pressure remains constant?


Question 6
The temperature inside my refrigerator is about 40 If I place a balloon in my fridge that initially has a temperature of 220 C and a volume of 0.5 liters, what will be the volume of the balloon in liters when it is fully cooled by my refrigerator?


Question 7
Some students believe that teachers are full of hot air. If I inhale 2.2 liters of gas at a temperature of 180 C and it heats to a temperature of 380 C in my lungs, what is the new volume of the gas in liters?


Question 8
Today, I forgot my soda in the trunk of my car. The initial pressure is 3 atm and it was a cool morning, at 15o By the afternoon, however, the temperature rose to 25oC. What is the pressure in atm inside the can?


please help me, im failing all my classes and really need some help with this. if i could give more than 100 i would

These questions all involve special cases of the ideal gas law, namely Boyle's, Charles', and Gay-Lussac's Laws. The ideal gas law relates together the absolute pressure (P), volume (V), the absolute temperature (T), and number of moles (n) of a gas by the following:

PV = nRT

where R is the universal gas constant.

The special cases of the ideal gas law are obtained by holding constant all but two of the variables of a gas.

Boyle's Law relates the pressure and volume of a given mass of gas at a constant temperature: PV = k or P₁V₁ = P₂V₂.

Charles' Law relates the volume and temperature of a given mass of gas at a constant pressure: V/T = k or V₁/T₁ = V₂/T₂.

Gay-Lussac's Law relates the pressure and temperature of a given mass of gas at a constant volume: P/T = k or P₁/T₁ = P₂/T₂.

Depending on what we're given and instructed to find in each question, we can figure out which law to use.

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Question 2:

We are given the volume of a gas at some pressure, and we're to find the new volume of the gas at a different pressure. Here, we use Boyle's Law: P₁V₁ = P₂V₂ where P₁ = 60 atm, V₁ = 20.0 L, and P₂ = 30 atm. We want to find V₂, which we can determine by rearranging the equation into the form V₂ = P₁V₁/P₂. Note that pressure and volume are inversely related according to Boyle's Law; since we're decreasing the pressure, the new volume of the gas should be greater than the initial volume of 20.0 L.

V₂ = (60 atm)(20.0 L)/(30.0 atm) = 40.0 L.

So, at 30 atm, the balloon will have a volume of 40.0 L.

---

Question 3:

This is another Boyle's Law question. The standard pressure (our initial pressure) is 1 atm. Here, we are decreasing the volume of the gas, and we want to find the new pressure; the pressure of the gas should thus increase proportionally (the pressure will be greater than 1 atm). Rearranging Boyle's Law to solve for P₂, we get P₂ = P₁V₁/V₂.

P₂ = (1 atm)(8.00 L)/(3 L) = 2.67 atm.

So, the new pressure of the gas is 2.67 atm (or 3 atm if we're considering V₂ to comprise one significant figure).

---

Question 4:

Here, we are increasing the temperature of a gas at a known pressure, and we want to determine what the new pressure will be. This is a Gay-Lussac's Law question; from the law, we see that pressure and temperature are directly proportional. Since we're increasing the temperature of the gas, we should expect the pressure of the gas to be greater than the initial 200 atm. Gay-Lussac's Law rearranged to solve for P₂ gives us P₂ = P₁T₂/T₁. When working with gas laws, temperatures must be in Kelvin (°C + 273.15 = K). So, T₁ = 300.15 K, T₂ = 350.15 K, and P₁ = 200 atm.

P₂ = (200 atm)(350.15 K)/(300.15 K) = 233 atm.

So, if the temperature is increased from 27 to 77 °C, the pressure of the gas in the tennis ball will be 233 atm. Here, it's ambiguous how many sig figs to use; if we use one sig fig per P₁, then our P₂ would equal P₁, which I think would be an absurd for a question to ask for. I would stick with either 233 atm or 230 atm (following the two sig figs of the temperatures), or you may go with however you've been instructed.

---

Question 5:

This is a Charles' Law question; we're looking for the new volume of a gas when the temperature of the gas is increased. As was the case in Gay-Lussac's Law, the two parameters in Charles' Law—volume and temperature—are directly proportional. Since the temperature of the gas is increased, we should expect the new volume of the gas to also increase (V₂ will be greater than 5.00 L). Temperatures should be in Kelvin.

V₂ = V₁T₂/T₁ = (5.00 L)(300.15 K)/(250.15 K) = 5.99 L.

---

Question 6:

Another Charles' Law question. As with question 5, we want to find the new volume of the gas after a change in temperature. This time, the final temperature is lower than the initial temperature, so we should expect that V₂ will be less than the initial 0.5 L. Again, temperatures in Kelvin.

V₂ = V₁T₂/T₁ = (0.5 L)(313.15 K)/(493.15 K) = 0.317 L.

So, the volume of the balloon when it is fully cooled by your refrigerator will be 0.317 L.

---

Question 7:

This is yet another Charles' Law question, and, again, we are solving for V₂ after a change in temperature. Since the final temperature is greater than the initial temperature, V₂ should be greater than 2.2 L. Again, the temperatures should be in Kelvin.

V₂ = V₁T₂/T₁ = (2.2 L)(653.15 K)/(453.15 K) = 3.17 L.

The new volume of the gas is 3.17 L ≈ 3.2 L (two sig figs).

---

Question 8:

We return to Gay-Lussac's Law here; pressure and temperature are directly proportional, and the temperature of the gas is increased. Thus, P₂ should be greater than 3 atm. Again, remember that temperatures must be in Kelvin.

P₂ = P₁T₂/T₁ = (3 atm)(298.15 K)/(288.15 K) = 3.1 atm.

So, the pressure inside the can after the temperature rise is 3.1 atm. Not a big increase, but an increase nonetheless.

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assaaap

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