A tank of volume V = 2.0 l will burst if the pressure exceeds 100 atm. The tank is filled with 6.0 mole of an ideal gas at a temperature of 23 ° C. We heat the tank.
a)What is the initial pressure in the tank?
b)At what temperature we can heat the tank with content without the burst?
T= 406 K = 133 C
To answer 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 the gas
R is the ideal gas constant
T is the temperature of the gas in Kelvin
Let's solve the questions step by step:
a) What is the initial pressure in the tank?
To find the initial pressure, we need to convert the temperature from Celsius to Kelvin and then use the ideal gas law equation.
We have:
V = 2.0 L
n = 6.0 mol
T = 23 °C + 273.15 = 296.15 K (converting Celsius to Kelvin)
R = 0.0821 L · atm/(mol · K) (ideal gas constant)
Now we can rearrange the ideal gas law equation to solve for P:
P = (nRT) / V
Substituting the given values into the equation:
P = (6.0 mol * 0.0821 L · atm/(mol · K) * 296.15 K) / 2.0 L
Calculating:
P = 738.559 atm
Therefore, the initial pressure in the tank is approximately 738.6 atm.
b) At what temperature can we heat the tank with its content without bursting?
We know that the tank will burst if the pressure exceeds 100 atm. So, we need to find the temperature at which the pressure equals 100 atm.
Again, we can rearrange the ideal gas law equation to solve for T:
T = (PV) / (nR)
Substituting the given values into the equation and solving for T:
T = (100 atm * 2.0 L) / (6.0 mol * 0.0821 L · atm/(mol · K))
Calculating:
T ≈ 33,333.33 K
Therefore, you would need to heat the tank to a temperature above approximately 33,333.33 K to cause it to burst.