A 0.63 mole sample of CO2 is in a 750 mL container. What is the pressure of the gas in torr when the has is held at -35.6oC?

12426torr

Explanation:

The following data were obtained from the question:

n = 0.63 mole

V = 750mL = 750/1000 = 0.75L

T = -35.6°C = -35.6 + 273 = 237.4K

R =0.082atm.L/Kmol

P =?

Using the ideal gas equation PV = nRT, the pressure can be obtained as follows:

PV = nRT

P = nRT/V

P = (0.63 x 0.082 x 237.4)/0.75

P = 16.35atm

Now let us convert this pressure (i.e 16.35atm) to a pressure in torr. This is illustrated below:

1atm = 760torr

16.35atm = 16.35 x 760 = 12426torr

Therefore, the pressure of the gas is 12426torr

The pressure will be 12441.2 torr

Explanation:

Step 1: Data given

Number of moles CO2 = 0.63 moles

Volume = 750 mL = 0.750 L

Temperature = -35.6 °C = 237.55 K

Step 2: Calculate pressure of the gas

p*V = n *R*T

⇒with p = the pressure of the gas = TO BE DETERMINED

⇒with V = the volume = 750 mL = 0.750 L

⇒with n = the number of moles CO2 = 0.63 moles

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

⇒with T = the temperature = 237.55 K

p = (n*R*T)/V

p = (0.63 * 0.08206  * 237.55) / 0.750

p = 16.37 atm = 12441.2 torr

The pressure will be 12441.2 torr