You transfer a sample of gas at 17 degrees celsius for a volume of 5.86 L and 1.10 atm to a container at 37 degrees celsius that has a pressure of 1.10 atm. what is the new volume of gas?
The combined gas law:
P2V2/T2=P1V1/T1 temps in Kelvins.
solve for V2
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin
First, let's convert the temperatures from Celsius to Kelvin:
Given initial temperature = 17 degrees Celsius
Initial temperature in Kelvin = 17 + 273 = 290 K
Given final temperature = 37 degrees Celsius
Final temperature in Kelvin = 37 + 273 = 310 K
Now, we can set up the equation using the initial conditions:
Initial pressure (P1) = 1.10 atm
Initial volume (V1) = 5.86 L
Initial temperature (T1) = 290 K
Next, we can use the same equation for the final conditions:
Final pressure (P2) = 1.10 atm
Final volume (V2) = ?
Final temperature (T2) = 310 K
Since the number of moles of gas (n) is the same before and after the transfer, we can cancel it out of the equation.
So, we can set up the following equation:
(P1 * V1) / T1 = (P2 * V2) / T2
Plugging in the given values:
(1.10 atm * 5.86 L) / 290 K = (1.10 atm * V2) / 310 K
Simplifying the equation:
V2 = [(1.10 atm * 5.86 L * 310 K) / (290 K * 1.10 atm)]
V2 = (1.10 * 5.86 * 310) / 290
V2 ≈ 6.187 L (rounded to three decimal places)
Therefore, the new volume of the gas is approximately 6.187 L.