I'm stuck on this assignment, +20 PTS and BRAINLIEST for step by step:

Assume constant n and T for questions 2; P1V1 = P2V2

2. If the pressure exerted on the gas in a 3.05-L balloon changes from 870 kPa to 969 kPa, what will be the new volume of the balloon?

Assume constant n and P for questions 3 & 4; V1/T1 = V2/T2

3. A sample of neon gas occupies 3.32 L at 360K. At what temperature would the volume of this sample of neon be at 4.50 L?

4. If 5.10 L of an ideal gas is cooled from -56°C to -82°C, what will be the new volume of the gas?

Assume constant P and T for questions 5 & 6: V1/n1 = V2/n2

5. If 0.750 mol of an ideal gas occupies a volume of 37.2 L at a particular temperature and pressure, what volume will 0.778 mol of an ideal gas occupy under the same conditions?

6. If 2.65 mol of an argon gas occupies a volume of 86.0 L at a particular temperature and pressure, how many moles would occupy a volume of 92.5 L under the same conditions?

Assume constant n and V for questions 7 & 8; P1/T1 = P2/T2

7. A sample of air exerts a pressure of 3.00 atm at 390K. To what temperature must the air be cooled so that it exerts a pressure of 2.45 atm?

8. A sample of helium in a rigid container exerts a pressure of 3.00 kPa at -12.3°C. What pressure will the helium exert if the temperature is changed to 25°C?

2. 2.74 L

3. 488 K

4. 7.47 L

5. 38.6 L

6. 2.85 mol

7. 319 K

8. 3.43 kPa

Explanation:

Generally speaking, you want to convert units to SI units, but in this case, we are working with ratios.  This makes up for using the units that wouldn't appropriate elsewhere.

2.  Use the equation P₁V₁ = P₂V₂.  Solve for V₂.  

(3.05 L)(870 kPa) = (969 kPa)(V₂)  

V₂ = 2.74 L.  

3.  Use the equation V₁/T₁ = V₂/T₂. Solve for T₂.  

(3.32 L)/(360 K) = (4.50 L)/(T₂)  

T₂ = 488 K.

4.  Do the same as above, but for V₂.  

(5.10 L)/(-56°C) = V₂/(-82°C)  

V₂ = 7.47 L

5.  Use the equation V₁/n₁ = V₂/n₂.  Solve for V₂.  

(37.2 L)/(0.750 mol) = (V₂)/(0.778 mol)  

V₂ = 38.6 L

6.  Do the same as above, but for n₂.

(86.0 L)/(2.65 mol) = (92.5 L)(n₂)  

n₂ = 2.85 mol

7.  Use the equation P₁/T₁ = P₂/T₂.  Solve for T₂.  

(3.00 atm)/(390 K) = (2.45 atm)/(T₂)  

T₂ = 319 K

8.  Do the same as above, but for P₂.  

In this specific case, however you will need to convert units.  Since both temperatures don't have the same sign, the ratio won't come out right.  Convert to Kelvin.  Add 273.15 to the temperature in Celsius to convert to Kelvin -12.3°C = 260.85 K  25°C = 298.15 K.

(3.00 kPa)/(260.85 K) = P₂/(298.15 K)

P₂ = 3.43 kPa

There is a lot in here... If you are confused about something, let me know!

the dependant variable is spilled water

the independent is the heat