Answer:
The volume of the tank is 23.28 m^3.
The final pressure of the mixture is 357.14 kPa.
Explanation:
Total moles of gas mixture (n) = 0.5 kmol Ar + 2 kmol N2 = 2.5 kmol
From the ideal gas equation, V = nRT/P
V is volume of the tank
n is the total moles of gas mixture = 2.5 kmol
R is gas constant = 8314.34 J/kmol.K
T is temperature = 280 K
P is pressure = 250 kPa = 250×1000 = 250,000 Pa
V = 2.5×8314.34×280/250,000 = 23.28 m^3
From Pressure law:
P1/T1 = P2/T2
P2 = P1T2/T1
P1 (initial pressure) = 250 kPa
T1 (initial temperature) = 280 K
T2 (final temperature) = 400 K
P2 (final pressure) = 250×400/280 = 357.14 kPa
Answer:
Explanation:
Given:
Moles of Argon, na = 0.5 kmol
Moles of N2, nn = 2 kmol
Pressure, P = 250 kPa
= 2.5 × 10^5 Pa
Temperature, T1 = 280 K
Final temperature, T2 = 400 K
Total number of moles in the system, n = na + nn
= 2 + 0.5
= 2.5 kmol
= 2500 mol.
Using ideal gas equation,
PV = nRT
Volume, V = (2500 × 8.314 × 280)/2.5 × 10^5
= 23.28 m^3
B.
Final pressure, P2 using pressure law,
P1/T1 = P2/T2
P2 = (P1 × T2)/T1
= (2.5 × 10^5 × 400)/280
= 3.57 × 10^5 Pa
= 357 kPa
