Given four cylinders containing equal amounts of oxygen gas, in which of the following would the oxygen molecules
collide most frequently?
a cylinder at a temperature of 268 k a volume of 0.5l
cylinder at a temperature of 298 k and a volume of 3,5l
cylinder at a temperature of 298 k and a volume of 0.5l
a cylinder at a temperature of 298 k and a volume of 2.5l

             Option-C (Cylinder at a temperature of 298 K and a volume of 0.5L) is the correct answer.

Explanation:

                   As we know that the gas molecules are free to move and collide with each other and with the walls of the container to exert pressure. In order to easily understand this problem we will assume that the gas exerting greater pressure will be having most frequent collisions with each other and with the walls of the container. Therefore, we will calculate pressure for each option using ideal gas equation assuming that the gas is acting as an ideal gas.

For Option 1:

                    Temperature = T = 268 K

                    Volume = V = 0.5 L

                    Moles  = n = 1    (which will remain constant)

                    Gas Constant = R = 0.0821 atm.L.mol⁻¹.K⁻¹

Formula used:

                    P V = n R T (Ideal Gas Equation)

Solving for P,

                    P = n R T / V

Putting Values,

                    P = 1 mol × 0.0821 atm.L.mol⁻¹.K⁻¹ × 268 K / 0.5 L

                   P₁ = 44.00 atm

For Option 2:

                    Temperature = T = 298 K

                    Volume = V = 3.5 L

                    Moles  = n = 1    

                    Gas Constant = R = 0.0821 atm.L.mol⁻¹.K⁻¹

Formula used:

                    P V = n R T

Solving for P,

                    P = n R T / V

Putting Values,

                    P = 1 mol × 0.0821 atm.L.mol⁻¹.K⁻¹ × 298 K / 3.5 L

                   P₂ = 6.99 atm

For Option 3:

                    Temperature = T = 298 K

                    Volume = V = 0.5 L

                    Moles  = n = 1  

                    Gas Constant = R = 0.0821 atm.L.mol⁻¹.K⁻¹

Formula used:

                    P V = n R T

Solving for P,

                    P = n R T / V

Putting Values,

                    P = 1 mol × 0.0821 atm.L.mol⁻¹.K⁻¹ × 298 K / 0.5 L

                   P₁ = 48.93 atm

For Option 4:

                    Temperature = T = 298 K

                    Volume = V = 2.5 L

                    Moles  = n = 1  

                    Gas Constant = R = 0.0821 atm.L.mol⁻¹.K⁻¹

Formula used:

                    P V = n R T

Solving for P,

                    P = n R T / V

Putting Values,

                    P = 1 mol × 0.0821 atm.L.mol⁻¹.K⁻¹ × 298 K / 2.5 L

                   P₁ = 9.78 atm

Result:

           Hence, it is proved that greater collisions occuring at 298 K when the volume was kept 0.5 L resulted in exerting the greatest pressure of 48.93 atm.

samuel de champlain. in 1608, king henry iv sponsored pierre dugua, sieur de mons and samuel de champlain as founders of the city of quebec with 28 men. this was the second permanent french settlement in the colony of canada.

out of these options:

ability to burn paper

pure sodium reacting violently with water

the boiling point of water

helium does not tend to react with anything

i would say this is a physical property:

helium does not tend to react with anything

why? :

i said helium does not tend to react with anything because well, it reacts to ! but ability to burn paper, pure sodium reacting violently with water, & the boiling point of water are chemical or

physical properties: the properties of a substance that can be observed without changing the chemical makeup of the substance

chemical properties: properties of a substance relating to the chemical nature and reactivity of the substance