`83Bi^(2H)(t 1/2= 130 sec)` decays to `81Tl^(207)` by `alpha`-emission. In an experiment starting with 5 moles of `83Bi^(211)`. how much pressure would be developed in a 350 L closed vessel at 25 C after 760 sec? [Antilog (1.759) = 57 41]

Options

A. 0.68 atm
B. 0.22 atm
C. 0.34 atm
D. 0.54 atm

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the decay process
The decay process is given as:
\[ ^{211}_{83}Bi \rightarrow ^{207}_{81}Tl + ^{4}_{2}He \]
This indicates that Bismuth-211 decays to Thallium-207 and emits an alpha particle (Helium-4).

### Step 2: Determine the decay constant (k)
The half-life (\( t_{1/2} \)) of Bismuth-211 is given as 130 seconds. The decay constant (\( k \)) can be calculated using the formula:
\[ k = \frac{\ln(2)}{t_{1/2}} \]
Substituting the values:
\[ k = \frac{0.693}{130} \approx 0.005338 \, \text{s}^{-1} \]

### Step 3: Calculate the number of moles of Bismuth remaining after 760 seconds
Using the formula for radioactive decay:
\[ N = N_0 e^{-kt} \]
Where:
- \( N_0 \) = initial number of moles of Bismuth = 5 moles
- \( k \) = decay constant = 0.005338 s\(^{-1}\)
- \( t \) = time = 760 seconds

Calculating \( N \):
\[ N = 5 e^{-0.005338 \times 760} \]
Calculating the exponent:
\[ -kt = -0.005338 \times 760 \approx -4.05288 \]
Now, calculating \( e^{-4.05288} \):
\[ N \approx 5 \times e^{-4.05288} \approx 5 \times 0.0176 \approx 0.088 \]
Thus, the remaining moles of Bismuth:
\[ N \approx 5 - 0.088 \approx 4.912 \]

### Step 4: Calculate the number of moles of Thallium produced
Since each decay of Bismuth produces one mole of Thallium:
\[ \text{Moles of } Tl = 0.088 \]

### Step 5: Calculate the pressure in the closed vessel
Using the ideal gas law:
\[ PV = nRT \]
Where:
- \( P \) = pressure (atm)
- \( V \) = volume (350 L)
- \( n \) = moles of gas (remaining Bismuth)
- \( R \) = ideal gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = temperature in Kelvin = 25°C = 298 K

Substituting the values:
\[ P \times 350 = 4.912 \times 0.0821 \times 298 \]
Calculating the right side:
\[ P \times 350 = 120.052 \]
Now solving for \( P \):
\[ P = \frac{120.052}{350} \approx 0.343 \]

### Final Answer
The pressure developed in the closed vessel after 760 seconds is approximately:
\[ P \approx 0.34 \, \text{atm} \]