Answer :
To solve this problem, we need to use the ideal gas law and the concept of mass conservation. The ideal gas law is given by [tex]PV = nRT[/tex], where [tex]P[/tex] is the pressure, [tex]V[/tex] is the volume, [tex]n[/tex] is the number of moles, [tex]R[/tex] is the ideal gas constant, and [tex]T[/tex] is the temperature in Kelvin.
Initial Conditions
- Initial mass of helium ([tex]m_1[/tex]): 9 kg
- Initial temperature ([tex]T_1[/tex]): 232°C = 505K (since [tex]T(K) = T(°C) + 273.15[/tex])
- Initial pressure ([tex]P_1[/tex]): 625 kPa
Sixty percent of the mass leaves the tank, leaving 40% of the original mass:
- Final mass of helium ([tex]m_2[/tex]): [tex]0.4 \times 9[/tex] kg = 3.6 kg
Final Conditions
- Final pressure ([tex]P_2[/tex]): 235 kPa
- **Final temperature ([tex]T_2[/tex]): ) ? (in Kelvin)
Since the tank's volume is constant and there is a mass change, let's use the relation of the ideal gas equation in terms of moles. The number of moles [tex]n[/tex] is given by [tex]n = \frac{m}{M}[/tex], where [tex]m[/tex] is the mass and [tex]M[/tex] is the molar mass of helium (approximately 4 kg/mol).
For initial conditions, the number of moles is:
[tex]n_1 = \frac{9 \text{ kg}}{4 \text{ kg/mol}} = 2.25 \text{ mol}[/tex]
For final conditions, the number of moles is:
[tex]n_2 = \frac{3.6 \text{ kg}}{4 \text{ kg/mol}} = 0.9 \text{ mol}[/tex]
Using the ideal gas law for initial conditions:
[tex]P_1V = n_1RT_1 \\
625 \times V = 2.25 \times R \times 505[/tex]
For final conditions:
[tex]P_2V = n_2RT_2 \\
235 \times V = 0.9 \times R \times T_2[/tex]
By dividing the two equations, [tex]V[/tex] and [tex]R[/tex] cancel out:
[tex]\frac{625}{235} = \frac{2.25 \times 505}{0.9 \times T_2}[/tex]
Solving for [tex]T_2[/tex]:
[tex]T_2 = \frac{0.9 \times 505 \times 235}{625 \times 2.25} \\
T_2 \approx 474.85 \text{ K}[/tex]
Convert [tex]T_2[/tex] from Kelvin to Celsius:
[tex]T_2(°C) = 474.85 - 273.15 \approx 201.7°C[/tex]
Final Answer
Based on the choices given, the closest option is VII. 201.7
Therefore, the final temperature in °C is 201.7.
