PLEASE HELP ASAP please answer both of these with work shown!!

What pressure is exerted by 0.450 mol of a gas at 25°C if the gas is in a 0.650-L container?

The recommended air pressure of a car tire is 32.0 psi (221 kPa). An average tire holds
around 1 mole of gas. What is the volume of the tire at 32°C?

The answer to your question is below

Explanation:

1.-

Data

Pressure = ?

moles = 0.450

temperature = 25°C

Volume = 0.650 L

R = 0.082 atm L / mol °K

1) Convert temperature to °K

°K = 25 + 273

    = 298°K

2) Use the Ideal gas law to solve this problem

     PV = nRT

- Solve for P

     P = nRT / V

- Substitution

      P = (0.45)(0.082)(298) / 0.65

- Result

      P = 16.91 atm

2.-

Pressure = 221 kPa

moles = 1

Volume = ?

Temperature = 32°C

R = 0.082 atm L /mol°K

Process

1) Convert pressure to atm

Pressure = 221 kPa = 221000 Pa

                   1 atm 101325 Pa

                    x       221000 Pa

                    x = (221000 x 1)/101325

                    x = 2.18 atm

2.- Convert temperature to °K

°K = 273 + 32 = 305°

3.- Substitution

     PV = nRT

- Solve for V

     V = nRT / P

- Substitution

     V = (1)(0.082)(305) / 2.18

- Result

      V = 11.47 L      

1) Pressure = 16.9 atm

2) The volume needed is 11.5 L

Explanation:

What pressure is exerted by 0.450 mol of a gas at 25°C if the gas is in a 0.650-L container?

p*V = n*R*T

⇒with p = the pressure = TO BE DETERMINED

⇒with V = the volume of the container = 0.650 L

⇒with n =the number of moles of gas = 0.450 moles

⇒ with R = the gas constant = 0.08206 L*atm /mol*K

⇒ with T = the temperature = 25 °C = 298 K

p = (n*R*T)/V

p = (0.450 * 0.08206 * 298) / 0.650

p = 16.9 atm

The recommended air pressure of a car tire is 32.0 psi (221 kPa). An average tire holds

around 1 mole of gas. What is the volume of the tire at 32°C?

p*V = n*R*T

⇒with p = the pressure = 32.0 psi = 221 kPa = 2.18 atm

⇒with V = the volume of the tire = TO BE DETERMINED

⇒with n =the number of moles of gas = 1.0 mol

⇒ with R = the gas constant = 0.08206 L*atm /mol*K

⇒ with T = the temperature = 32 °C = 305 K

V = (n*R*T) / p

V = 1.0 * 0.08206 * 305) / 2.18

V = 11.5 L