Nitrous oxide (N2O) is used as an anesthetic. The pressure on 2.50 L of N2O changes from 105kPa to 40.5 kPa. If the temperature does not change, what will the new volume be?
I know the Boyle's law is
P1 x V1 = P2 x V2
You have the law correct, you are looking for V2
V2=P1*V1/P2. Pressure was reduced by a factor of 2.5 (about) so volume ought to be increased by that same factor.
nitrous oxide ( n 2 o ) is used as an anesthetic the pressureon 2.50 L of
(n 2 o ) changes from 105 kpa to 40.5
To find the new volume of the N2O, we can use Boyle's Law, which states that the product of the initial pressure and volume is equal to the product of the final pressure and volume.
Boyle's Law formula: P1 x V1 = P2 x V2
Given:
Initial pressure (P1) = 105 kPa
Initial volume (V1) = 2.50 L
Final pressure (P2) = 40.5 kPa
Let's plug in the given values into Boyle's Law formula to find the new volume (V2):
P1 x V1 = P2 x V2
105 kPa x 2.50 L = 40.5 kPa x V2
Dividing both sides of the equation by 40.5 kPa:
(105 kPa x 2.50 L) / 40.5 kPa = V2
Calculating:
(262.5 kPa * L) / 40.5 kPa = V2
6.481 L = V2
Therefore, the new volume of N2O will be 6.481 L when the pressure changes from 105 kPa to 40.5 kPa, assuming the temperature remains constant.
To find the new volume using Boyle's Law, you need to rearrange the formula:
\(P_1 \times V_1 = P_2 \times V_2\)
Given:
\(P_1 = 105 \, \text{kPa}\)
\(V_1 = 2.50 \, \text{L}\)
\(P_2 = 40.5 \, \text{kPa}\)
\(V_2 = ?\)
Substituting the given values into the formula:
\(105 \, \text{kPa} \times 2.50 \, \text{L} = 40.5 \, \text{kPa} \times V_2\)
Now, isolate \(V_2\) by dividing both sides of the equation by \(40.5 \, \text{kPa}\):
\(\frac{105 \, \text{kPa} \times 2.50 \, \text{L}}{40.5 \, \text{kPa}} = V_2\)
Simplifying the equation:
\(V_2 = 6.519 \, \text{L}\)
Therefore, the new volume will be approximately 6.519 liters when the pressure changes from 105 kPa to 40.5 kPa, assuming the temperature remains constant.