Answer:
Pressure of gas A = 544.324 kPa
Pressure of gas B = 2217.784 kPa
Explanation:
Data provided in the question:
Total volume of the cylinder, V = 5.55 L = 0.00555 m³ [1 m³ = 1000 L]
Moles on gas A, [tex]n_a[/tex] = 1.21 mol
Moles on gas A, [tex]n_b[/tex] = 4.93 mol
Temperature, T = 27.3°C = 27.3 + 273 = 300.3 K
Now,
Pressure = [tex]\frac{nRT}{V}[/tex]
here,
R is the ideal gas constant = 8.314 J/mol.K
Therefore,
Pressure of gas A = [tex]\frac{n_aRT}{V}[/tex]
= [tex]\frac{1.21\times8.314\times300.3}{0.00555}[/tex]
= 544324.32 Pa
= 544.324 kPa
Pressure of gas B = [tex]\frac{n_bRT}{V}[/tex]
= [tex]\frac{4.93\times8.314\times300.3}{0.00555}[/tex]
= 2217784.22 Pa
= 2217.784 kPa
