Respuesta :
Answer:
(a) 62.57 L
(b) 801.94 kPa
Explanation:
Given:
[tex]n[/tex] = number of moles of gas = 10.3 mol
[tex]P_1[/tex] = initial pressure of the gas = [tex]519\ kPa = 5.19\times 10^5\ Pa[/tex]
[tex]T_1[/tex] = initial temperature of the gas = [tex]361.38^\circ C = (361.38+273)\K = 598.38\ K[/tex]
[tex]T_2[/tex] = final temperature of the gas = [tex]651.6^\circ C = (651.6+273)\K = 924.6\ K[/tex]
[tex]V[/tex] = volume of the tank
R = universal gas constant = [tex]8.314 J/mol K[/tex]
Part (a):
Using Ideal gas equation, we have
[tex]PV=nRT\\\Rightarrow V = \dfrac{nRT}{P}\\\Rightarrow V = \dfrac{10.3\times 8.314\times 598.38}{5.19\times 10^{5}}\\\Rightarrow V = 6.2567\times10^{-2}\ m^3\\\Rightarrow V = 62.567\ L[/tex]
Hence, the volume of the container is 62.567 L.
Part (b):
As the volume of the container remains constant.
Again using ideal gas equation,
[tex]PV=nRT\\\because V,\ n, R\ are\ constant\\\therefore \dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}\\\Rightarrow P_2 = \dfrac{T_2}{T_1}P_1\\\Rightarrow P_2 = \dfrac{924.6}{598.38}\times 519\\\Rightarrow P_2 = 801.94\ kPa[/tex]
Hence, the final pressure of the gas is now 801.94 kPa.
