Gas is confined in a tank at a pressure of 12.7 atm and a temperature of 22.0°C. If two-thirds of the gas is withdrawn and the temperature is raised to 79.0°C, what is the pressure of the gas remaining in the tank?
P V = n R T
To find the pressure of the gas remaining in the tank, we can use the combined gas law, which relates the initial and final conditions of a gas.
The combined gas law equation is given as:
(P1 × V1) / (T1) = (P2 × V2) / (T2)
Where:
P1 = initial pressure of the gas
V1 = initial volume of the gas
T1 = initial temperature of the gas
P2 = final pressure of the gas (what we are trying to find)
V2 = final volume of the gas (since two-thirds is withdrawn, the final volume will be 1/3 of the initial volume)
T2 = final temperature of the gas
Let's substitute the given values into the equation:
P1 = 12.7 atm (given)
V1 = initial volume (unknown)
T1 = 22.0°C (given, but needs to be converted to Kelvin)
P2 = final pressure (what we are trying to find)
V2 = (1/3) × V1 (two-thirds of the gas is withdrawn, so the final volume is 1/3 of the initial volume)
T2 = 79.0°C (given, but needs to be converted to Kelvin)
First, we need to convert the temperatures from Celsius to Kelvin by adding 273.15:
T1 = 22.0°C + 273.15 = 295.15 K
T2 = 79.0°C + 273.15 = 352.15 K
Now, we need to find the initial volume, V1. Unfortunately, it is not given in the question. Without the value of V1, we cannot calculate the final pressure, P2.
Therefore, we cannot determine the pressure of the gas remaining in the tank without the initial volume.