Consider a cylindric container with the radius of circle 50 cm and the length of 3 m. We fill this container with one mole of the oxygen gas O2 at the room temperature 20°C. Assume it is an ideal gas.

a) What is the density of the gas in the tank?

b) What is the pressure of the gas?

c) We compress the gas by moving one of the walls and reducing the length to 1m while keeping the same radius of circle). What is the pressure of the gas if the temperature stays the same?

(a) The density of gas in the tank is 1.36×10^-5 g/cm^3

(b) The pressure of the gas is 0.0102 atm

(c) The pressure of the gas is 0.0306 atm

Explanation:

(a) Density = mass/volume

number of moles of O2 = 1 mol

MW of O2 = 32 g/mol

mass = number of moles × MW = 1 × 32 = 32 g

Volume of cylinder (V) = πr^2l

r is radius = 50 cm

l is length = 3m = 3×100 = 300 cm

V = 3.142 × 50^2 × 300 = 2,356,500 cm^3

Density = 32 g/2,356,500 cm^3 = 1.36×10^-5 g/cm^3

(b) From ideal gas equation

P = rho × R × T/MW

rho is density of the gas = 1.36×10^-5 g/cm^3

R is gas constant = 82.057 cm^3.atm/gmol.K

T is temperature = 20 °C = 20+273 = 293 K

MW is molecular weight of O2 = 32 g/mol

P = 1.36×10^-5 × 82.057 × 293/32 = 0.0102 atm

(c) New length = 1 m = 1×100 = 100 cm

Volume = πr^2l = 3.142 × 50^2 × 100 = 785,500 cm^3

Density = mass/volume = 32/785,500 = 4.07×10^-5 g/cm^3

P = rho × R × T/MW = 4.07×10^-5 × 82.057 × 293/32 = 0.0306 atm

x+2 because it just doubles hon so the pattern can actually tell you the expression.

so for example,

if x=2 then 2+2 is 4

if x=4 then 4+4 is 8

and so on and so forth.

i really hope that this makes sense and you out in a way hon: )