Question

A) A 193.0 L sample of gas of the upper atmosphere at a pressure of 4.8 torr is compressed into a 256.0 mL container at the same temperature. What is the new pressure?

B) To what volume would the original sample have had to be compressed to reach a pressure of 7.0 atm?

Answer 1

(A) We know that PV = nRT

Where

T = Temperature ;P = pressure ; n = No . of moles ;R = gas constant ; V= Volume of the gas

As the gas remains the same , n will constant also T kept constant

So PV / T = constant

Thereby PV = P'V'

Where

P = initial pressure = 4.8 torr

V = initial volume = 193.0 L

P'= final pressure = ?

V' = final volume = 256.0 mL = 0.256 L

Plug the values we get P' = (PV) / V'

                                     = ( 4.8x193) / 0.256

                                     = 3.6x10³ torr

(A) We know that PV = nRT

Where

T = Temperature ;P = pressure ; n = No . of moles ;R = gas constant ; V= Volume of the gas

As the gas remains the same , n will constant also T kept constant

So PV / T = constant

Thereby PV = P'V'

Where

P = initial pressure = 4.8 torr

V = initial volume = 193.0 L

P'= final pressure = 7.0 atm = 7.0x760 torr = 5320 torr

V' = final volume = ?

Plug the values we get V' = (PV) / P'

                                     = ( 4.8x193) / 5320

                                     = 0.174 L

                                     = 174 mL