Let's solve this problem step-by-step using the provided information and the Ideal Gas Law.
Given:
- Initial volume ([tex]\( V_1 \)[/tex]) = 32.1 L (Constant volume, hence [tex]\( V_1 = V_2 \)[/tex])
- Initial temperature ([tex]\( T_1 \)[/tex]) = [tex]\( 29^{\circ} C \)[/tex]
- Initial pressure ([tex]\( P_1 \)[/tex]) = 15.2 atm
- Final temperature ([tex]\( T_2 \)[/tex]) = [tex]\( 105^{\circ} C \)[/tex]
- Universal Gas Constant ([tex]\( R \)[/tex]) = 0.082 L atm / K mol (though not needed explicitly here)
First, convert the given temperatures from Celsius to Kelvin:
[tex]\[ T_1 = 29 + 273 = 302 \, \text{K} \][/tex]
[tex]\[ T_2 = 105 + 273 = 378 \, \text{K} \][/tex]
Since the volume remains constant and we are dealing with an ideal gas, we use the relationship derived from the Ideal Gas Law:
[tex]\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \][/tex]
Rearranging the equation to solve for the final pressure [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = P_1 \times \frac{T_2}{T_1} \][/tex]
Now, substitute the known values:
[tex]\[ P_2 = 15.2 \, \text{atm} \times \frac{378 \, \text{K}}{302 \, \text{K}} \][/tex]
Calculate [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = 15.2 \times 1.2516556291390728 \][/tex]
[tex]\[ P_2 \approx 19.025165562913905 \, \text{atm} \][/tex]
Rounding to one decimal place, the final pressure in the cylinder is:
[tex]\[ P_2 \approx 19.0 \, \text{atm} \][/tex]
So, the correct answer is:
[tex]\[ \boxed{19.0} \][/tex]
