A gas occupying 43 mL at standard conditions
is heated to 29 celsius while the pressure is reduced
to 0.84091 atm. What is the new volume
occupied by the gas?
(P1V1/T1) = (P2V2/T2)
Remember T must be in kelvin
To solve this problem, we can use the combined gas law, which relates the initial and final conditions of a gas sample. The combined gas law equation is:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
- P1 is the initial pressure (in atm)
- V1 is the initial volume (in mL)
- T1 is the initial temperature (in Celsius)
- P2 is the final pressure (in atm)
- V2 is the final volume (unknown)
- T2 is the final temperature (in Celsius)
Let's substitute the given values into the equation:
(P1 * V1) / T1 = (P2 * V2) / T2
Given:
P1 = 1 atm (standard conditions)
V1 = 43 mL
T1 = 273.15 K (standard conditions)
P2 = 0.84091 atm
T2 = 29 Celsius = 29 + 273.15 = 302.15 K
Plugging in the values, the equation becomes:
(1 atm * 43 mL) / 273.15 K = (0.84091 atm * V2) / 302.15 K
Substituting the known values into the equation, we have:
(43 mL) / 273.15 K = (0.84091 atm * V2) / 302.15 K
Simplifying, we get:
(43 * 302.15) / (0.84091 * 273.15) = V2
Solving this, we find:
V2 ≈ 57.505 mL
Therefore, the new volume occupied by the gas after heating and pressure reduction is approximately 57.505 mL.
To find the new volume occupied by the gas, we can use the combined gas law formula, which combines Boyle's law and Charles' law. The formula is:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where
P1 = initial pressure (in atm)
V1 = initial volume (in mL)
T1 = initial temperature (in Kelvin)
P2 = final pressure (in atm)
V2 = final volume (unknown)
T2 = final temperature (in Kelvin)
Let's convert the given values to the appropriate units before substituting them into the formula:
Initial volume (V1) = 43 mL
Final pressure (P2) = 0.84091 atm
Initial temperature (T1) = standard temperature in Celsius + 273.15 = 0°C + 273.15 = 273.15 K
Final temperature (T2) = 29°C + 273.15 = 302.15 K
Now we can substitute the values into the formula to solve for the final volume (V2):
(P1 * V1) / (T1) = (P2 * V2) / (T2)
(1 atm * 43 mL) / (273.15 K) = (0.84091 atm * V2) / (302.15 K)
43 / 273.15 = (0.84091 * V2) / 302.15
0.1574 = (0.84091 * V2) / 302.15
To solve for V2, we can rearrange the equation:
0.1574 * 302.15 = 0.84091 * V2
47.55841 = 0.84091 * V2
Divide both sides by 0.84091:
47.55841 / 0.84091 = V2
V2 ≈ 56.562 mL
Therefore, the new volume occupied by the gas is approximately 56.562 mL.