What pressure is exerted by 0.90 moles of gas at 50 C if the gas is in a 1.25 L container?

19.07 atm

Explanation:

Data obtained from the question include the following:

Number of mole (n) = 0.9 mole

Temperature (T) = 50°C = 50°C + 273 = 323K

Volume (V) = 1.25L

Pressure (P) =...?

Note: the gas constant (R) = 0.082atm.L/Kmol

The pressure exerted by the gas can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

P = nRT /V

P= (0.9 x 0.082 x 323)/1.25

P = 19.07 atm

Therefore, the pressure exerted by the gas is 19.07 atm

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.