Asample of he gas (3.0 l) at 5.6 atm and 25°c was combined with 4.5 l of ne gas at 3.6 atm and 25°c at constant temperature in a 9.0 l flask. the total pressure in the flask was atm. assume the initial pressure in the flask was 0.00 atm.

3.6667

Explanation:

For helium gas:

Using Boyle's law  

Given ,  

V₁ = 3.0 L

V₂ = 9.0 L

P₁ = 5.6 atm

P₂ = ?

Using above equation as:

The pressure exerted by the helium gas in 9.0 L flask is 1.8667 atm

For Neon gas:

Using Boyle's law  

Given ,  

V₁ = 4.5 L

V₂ = 9.0 L

P₁ = 3.6 atm

P₂ = ?

Using above equation as:

The pressure exerted by the neon gas in 9.0 L flask is 1.8 atm

Thus total pressure = 1.8667 + 1.8 atm = 3.6667 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.