Gas is confined in a tank at a pressure of
10 atm and a temperature of 14◦C.
If 58 percent of the gas is withdrawn and
the temperature is raised to 50◦C, what is the new pressure in the tank?
To find the new pressure in the tank, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature
Given:
P1 = 10 atm (initial pressure)
T1 = 14◦C (initial temperature)
T2 = 50◦C (final temperature)
Withdrawn gas = 58%
First, let's convert the initial temperature from Celsius to Kelvin by adding 273.15:
T1 = 14 + 273.15 = 287.15 K
Next, let's convert the final temperature from Celsius to Kelvin:
T2 = 50 + 273.15 = 323.15 K
To find the new pressure, we need to find the new volume of the gas. Since the gas is withdrawn, the volume will not remain constant.
Let's assume the initial volume is V1, and the final volume is V2.
Since the gas withdrawn is 58%, the remaining gas is 42%:
Remaining gas = 100% - Withdrawn gas
Remaining gas = 100% - 58% = 42%
The volume is directly proportional to the number of moles of gas when pressure and temperature are constant. Therefore, we can write the following equation:
(V1/V2) = (n1/n2)
Let's assume we have 100 units of gas initially, so the remaining gas would be 42 units.
Therefore, (V1/V2) = (100/42)
Now, we need to find the change in volume by calculating the ratio of V1 to V2:
V2 = V1 / (100/42)
Now, we can substitute the values into the ideal gas law equation:
P1V1/T1 = P2V2/T2
Solving for P2, we get:
P2 = (P1V1T2) / (T1V2)
Substitute the given values into the equation:
P2 = (10 * V1 * 323.15) / (287.15 * V2)
Now, we can substitute the calculated value for V2 into the equation to find P2.
P2 = (10 * V1 * 323.15) / (287.15 * (V1 / (100/42)))
By simplifying the equation, we get:
P2 = 14.1052 * P1
Now, substitute the given value of P1 (10 atm) into the equation:
P2 = 14.1052 * 10
Therefore, the new pressure in the tank is approximately 141.05 atm.