PLEASE HELP ME WITH THESE CHEMISTRY PROBLEMS?!?!?STEP BY STEP PLEASE!!! IM SO CONFUSED?

In chemistry, you learn that the volume of one mole of an ideal gas a standard temperature and pressure is 22.4 liters. Standard temperature is 273˚K, and standard pressure is one atmosphere. The number of moles of the gas is not affected by the temperature and pressure. Let’s determine the volume of 34.5 moles of the gas at standard temperature and pressure.

V = 34.5 * 22.4 = 772.8 liters

Now we can use the following equation to determine the volume of the gas at a pressure of 2.23 atmospheres and 411˚ K.

P1 * V1 ÷ T1 = P2 * V2 ÷ T2

1 * 772.8 ÷ 273 = 2.23 * V2 ÷ 411

V2 * 608.79 = 317,620.8

V2 = 317,620.8 ÷ 608.79

The volume is approximately 552 liters. These are the two steps that I used to solve this type of problem. Many people use the following equation.

P * V = n * R * T

The exact value of R is 22.4 ÷ 273

2.23 * V = 34.5 * (22.4 ÷ 273) * 411

V = 317,620.8 ÷ 608.79

The volume is approximately 552 liters. You get the same answer.

A hot air balloon is filled with 1.41 × 106 L of an ideal gas on a cool morning (11 °C). The air is heated to 121 °C. What is the volume of the air in the balloon after it is heated? Assume that none of the gas escapes from the balloon.

To solve this problem, both of the temperatures must be converted from ˚C to ˚K by adding 273˚.

T1 = 11 + 273 = 284

T2 = 121 + 273 = 394

Now we can use these numbers in the following equation to determine the new volume.

V1 ÷ T1 = V2 ÷ T2

1.41 * 10^6 ÷ 284 = V2 ÷ 394

V2 = 5.5554 * 10^8 ÷ 284

The volume is approximately 1.96 * 10^6 liters. I hope this is helpful for you.

V = nRT / P = (34.5 mol) x (0.08205746 L atm/K mol) x (411 K) / (2.23 atm) = 522 L

(1.41 × 10^6 L) x (121 + 273) K / (11 + 273) K = 1.96 × 10^6 L

If 34.5 mol of an ideal gas is at 2.23 atm and 411 K, what is the volume of the gas

PV = nRT

V = [34.5 X 0.0821 X 411] / 2.23 <<< use a calculator.. answer will be L

A hot air balloon is filled with 1.41 × 106 L of an ideal gas on a cool morning (11 °C). The air is heated to 121 °C. What is the volume of the air in the balloon after it is heated? Assume that none of the gas escapes from the balloon.

1.41 × 106 L X [121 + 273] / [11 + 273] <<< use a calculator