A rigid tank contains 0.5 kmol of Ar and 2 kmol of N2 at 250 kPa and 280 K. The mixture is now heated to 400 K. Determine the volume of the tank and the final pressure of the mixture.

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

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

3.2x10^7