To find the original size of the container, we can use the ideal gas law:
\[ PV = nRT \]
where:
- \( P \) is the pressure in atm,
- \( V \) is the volume in liters,
- \( n \) is the number of moles of gas,
- \( R \) is the universal gas constant, \( 0.0821 , \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \),
- \( T \) is the temperature in Kelvin.
Step 1: Convert the initial temperature to Kelvin.
The initial temperature is given as \( 30.0^\circ C \).
\[ T_1 = 30.0 + 273.15 = 303.15 , K \]
Step 2: Find the initial volume using the initial conditions.
We can use the ideal gas law with the initial conditions:
\[ P_1 = 1.5 , \text{atm} \] \[ n_1 = 2.22 , \text{moles} \] \[ T_1 = 303.15 , K \]
Substituting these values into the ideal gas law:
\[ V_1 = \frac{nRT}{P} \]
\[ V_1 = \frac{(2.22 , \text{moles}) \times (0.0821 , \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}) \times (303.15 , K)}{1.5 , \text{atm}} \]
Calculating \( V_1 \):
\[ V_1 = \frac{(2.22) \times (0.0821) \times (303.15)}{1.5} \]
\[ V_1 = \frac{(54.3012)}{1.5} \approx 36.2008 , \text{L} \]
So, the original volume of the container is approximately \( 36.20 , \text{L} \).
Step 3: Verify with final conditions.
The final conditions are given as:
- Pressure increases by \( 0.500 , \text{atm} \) from \( 1.5 , \text{atm} \) to \( 2.0 , \text{atm} \) (final pressure \( P_2 = 2.0 , \text{atm} \)).
- Volume is \( 4.30 , L \).
- Amount is \( 3.45 , moles \).
- Temperature increases by \( 33.0^\circ C \) (final temperature \( T_2 = 30.0 + 33.0 = 63.0^\circ C = 336.15 , K \)).
Using the final conditions:
\[ V_2 = \frac{nRT}{P} \]
\[ V_2 = \frac{(3.45 , \text{moles}) \times (0.0821) \times (336.15)}{2.0} \]
Calculating \( V_2 \):
\[ V_2 = \frac{(3.45) \times (0.0821) \times (336.15)}{2.0} \]
\[ = \frac{(85.122517)}{2.0} \approx 42.5613 , \text{L} \]
This result doesn't match the expected volume. It's possible that the initial volume provided was different than the ideal gas law suggests or there might be other factors at play such as a leak or gas loss between the measurements.
Thus, the original size of the container based on the calculations for the initial conditions is:
Original Volume: \( 36.20 , \text{L} \).