A sample of argon at 2.0 atm pressure is heated until its temperature is raised from 60 °C to 90 °C. The pressure of the gas at this temperature is 3.0 atm and the volume is 4.4 L.
What was the initial volume of the gas?
(P1V1/T1) = (P2V2/T2)
Rember T must be in kelvin.
I have tried these calculations. I believe the answer is 6.1, is this correct?
I agree
To solve this problem, we can use the combined gas law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
In this equation:
P1 and P2 are the initial and final pressures of the gas,
V1 is the initial volume of the gas,
T1 and T2 are the initial and final temperatures of the gas,
and V2 is the final volume of the gas.
We are given:
P1 = 2.0 atm,
T1 = 60°C (which needs to be converted to Kelvin),
P2 = 3.0 atm,
T2 = 90°C (which needs to be converted to Kelvin),
and V2 = 4.4 L.
First, let's convert the temperatures to Kelvin by adding 273.15 to each Celsius temperature:
T1 = 60 + 273.15 = 333.15 K
T2 = 90 + 273.15 = 363.15 K
Now, we can substitute the given values into the combined gas law equation:
(2.0 atm * V1) / (333.15 K) = (3.0 atm * 4.4 L) / (363.15 K)
Next, we can solve for V1 (initial volume) by rearranging the equation:
V1 = (3.0 atm * 4.4 L * 333.15 K) / (2.0 atm * 363.15 K)
Calculating this expression will give us the initial volume of the gas.