1.given 30 m^3 of helium at 15 degree c and 1.0x10^5 Pa, determine the volume at -6 degree c and 5.6x10^4 Pa
2. if a mass of gas occupies 500 l at 16 degree c and 10^5 Pa,determine the volume at 38 degree c and 1.7x10^5 Pa.
3. Calculate the final temperature required to change 10 l of helium at 100 k and 10^4 Pa to 20 l at 2x10^4 Pa.
a. combined gas law
b. ibid
c. ibid
Do you have a question about how to use the combined gas law?
P1V1/T1=P2V2/T2 temps in kelvins
To solve these questions, we can use the ideal gas law equation, which states:
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, and
T is the temperature of the gas in Kelvin.
To solve for different variables in the ideal gas law equation, we need to rearrange the equation accordingly.
Now let's solve each question step by step.
1. Given 30 m^3 of helium at 15 degrees Celsius and 1.0x10^5 Pa, we need to determine the volume at -6 degrees Celsius and 5.6x10^4 Pa.
Step 1: Convert temperatures to Kelvin.
To convert Celsius to Kelvin, add 273.15. Therefore:
Initial temperature (TF) = 15 + 273.15 = 288.15 K
Final temperature (TI) = -6 + 273.15 = 267.15 K
Step 2: Rearrange the ideal gas law equation to solve for the final volume (VF).
PV / T = nR (as the number of moles and the gas constant remain constant)
Since P1V1 / T1 = P2V2 / T2 (Boyle's Law for constant number of moles)
P1V1 / T1 = P2V2 / T2
(1.0x10^5 Pa)(30 m^3) / (288.15 K) = (5.6x10^4 Pa)(VF) / (267.15 K)
Step 3: Solve for VF (Volume at -6 degrees Celsius and 5.6x10^4 Pa).
VF = [(1.0x10^5 Pa)(30 m^3)(267.15 K)] / [(5.6x10^4 Pa)(288.15 K)]
VF = 23.56 m^3
Therefore, the volume of helium at -6 degrees Celsius and 5.6x10^4 Pa would be approximately 23.56 m^3.
2. Given a gas mass that occupies 500 L at 16 degrees Celsius and 10^5 Pa, we need to determine the volume at 38 degrees Celsius and 1.7x10^5 Pa.
Step 1: Convert temperatures to Kelvin.
Initial temperature (TF) = 16 + 273.15 = 289.15 K
Final temperature (TI) = 38 + 273.15 = 311.15 K
Step 2: Rearrange the ideal gas law equation to solve for the final volume (VF).
P1V1 / T1 = P2V2 / T2
(10^5 Pa)(500 L) / (289.15 K) = (1.7x10^5 Pa)(VF) / (311.15 K)
Step 3: Solve for VF (Volume at 38 degrees Celsius and 1.7x10^5 Pa).
VF = [(10^5 Pa)(500 L)(311.15 K)] / [(1.7x10^5 Pa)(289.15 K)]
VF = 1028.43 L
Therefore, the volume of the gas at 38 degrees Celsius and 1.7x10^5 Pa would be approximately 1028.43 L.
3. Calculate the final temperature required to change 10 L of helium at 100 K and 10^4 Pa to 20 L at 2x10^4 Pa.
Step 1: Rearrange the ideal gas law equation to solve for the final temperature (TF).
P1V1 / T1 = P2V2 / T2
(10^4 Pa)(10 L) / (100 K) = (2x10^4 Pa)(20 L) / (TF)
Step 2: Solve for TF (Final temperature).
TF = [(10^4 Pa)(10 L)(2x10^4 Pa)(20 L)] / [(100 K)]
TF = 4x10^7 K
Therefore, the final temperature required to change 10 L of helium at 100 K and 10^4 Pa to 20 L at 2x10^4 Pa would be 4x10^7 K.