The nuclide 38cl decays by beta emission with a half life of 40min. A sample of .40 m of H38cl is placed in a 6.24 liter container. After 80 min the pressure is 1650 mmHg. What is temperature of container?
I have tried calculating the decay rate of the 38cl and then plugging into the formula pv -net but I do not get the answer, 300 k. Any help would be great.
To solve this problem, we need to first calculate the number of moles of H38Cl remaining in the container after 80 minutes, and then use the ideal gas law equation to find the temperature.
Step 1: Calculate the number of moles of H38Cl remaining
Given that the half-life of 38Cl is 40 minutes, we can use the half-life formula to calculate the fraction of the original material remaining after 80 minutes:
(1/2)^(t/half-life) = (1/2)^(80/40) = (1/2)^2 = 1/4
Therefore, after 80 minutes, only 1/4 of the initial H38Cl remains in the container.
Step 2: Calculate the number of moles of H38Cl
To calculate the number of moles of H38Cl remaining in the container, we need to know the initial number of moles present. We are given that the sample contains 0.40 mol of H38Cl initially.
Therefore, the number of moles remaining after 80 minutes is (1/4) * 0.40 mol, which is 0.10 mol.
Step 3: Use the ideal gas law to find the temperature
The ideal gas law equation is:
PV = nRT
Where:
P is the pressure (in atm)
V is the volume (in liters)
n is the number of moles
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature (in Kelvin)
Rearranging the equation to solve for T, we have:
T = PV / (nR)
Now we can substitute the given values into the equation:
P = 1650 mmHg = 1650/760 = 2.171 atm
V = 6.24 L
n = 0.10 mol
R = 0.0821 L·atm/(mol·K)
T = (2.171 atm * 6.24 L) / (0.10 mol * 0.0821 L·atm/(mol·K))
Canceling out units and evaluating the expression, we find:
T ≈ 300 K
Therefore, the temperature of the container is approximately 300 K.