Answer:
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.
