a metal tank with a volume of 3.10 L will burst if the absolute pressure of the gas exceeds 100 atm. If 11 mol of an ideal gas is put into the tank at a temperature of 23.0 C, to what temperature can the gas be warmed before the tank ruptures? Neglect the thermal expansion of the tank.
Does the PV=nRT formula work for this question?
yes, that will work, be certain to use kelvins
So pressure will be 100 atm.
Volume will be 3.10 L
Temperature will be 296 K
n will be 11 moles
R will be 0.0821 (L*atm)/(mol*K)
There is no unknown to solve for?
I am confused..
Temperature is unknown, you are looking for that.
T=PV/nR
Yes, the ideal gas law formula PV = nRT can be used to solve this question. The ideal gas law relates the pressure (P), volume (V), number of moles (n), temperature (T) in Kelvin, and the ideal gas constant (R). In this case, we need to determine the temperature (T) at which the gas will cause the tank to rupture.
To use the ideal gas law, we need to convert the given values to appropriate units. The volume is already given in liters (L), and the number of moles (n) is given as 11 mol. The temperature (T) in Celsius needs to be converted to Kelvin by adding 273.15. So, the given temperature of 23.0 °C becomes 23.0 + 273.15 = 296.15 K.
Now, we can rearrange the ideal gas law equation to solve for temperature (T):
PV = nRT
T = PV / (nR)
Substituting the given values:
T = (100 atm) * (3.10 L) / (11 mol * R)
Before proceeding further, let's identify the value of the ideal gas constant (R). The ideal gas constant (R) is approximately 0.0821 L · atm / (mol · K).
Now, we can substitute the value of R and perform the calculations:
T = (100 atm) * (3.10 L) / (11 mol * 0.0821 L · atm / (mol · K))
Simplifying:
T = 2681.8 atm·L / (90.31 mol·K)
Canceling units:
T ≈ 29.7 K
Therefore, the gas can be warmed to a temperature of approximately 29.7 K before the tank ruptures.