A sample of gas occupies 18 L under a pressure of 1.3 atm. What would the resulting valume be if the pressure were increased to 3.9 atm if the temperature did not change?
P*V would be constant in the process.
What volume (in liters) would the gas occupy at a pressure of 1.20 atm if its temperature remained unchanged?
To find the resulting volume when the pressure is increased, we can use the combined gas law, which relates the initial and final conditions of a gas sample. The combined gas law formula is:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature
In this case, we are told that the temperature does not change (T1 = T2), so we can simplify the equation to:
(P1 * V1) = (P2 * V2)
Plugging in the given values:
P1 = 1.3 atm
V1 = 18 L
P2 = 3.9 atm
Now we can solve for V2:
(1.3 atm * 18 L) = (3.9 atm * V2)
Rearranging the equation to solve for V2:
V2 = (1.3 atm * 18 L) / 3.9 atm
V2 ≈ 5.999 L
Therefore, the resulting volume would be approximately 5.999 L if the pressure were increased to 3.9 atm while keeping the temperature constant.