## 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

Question

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.

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3 years 2021-08-02T14:51:09+00:00 2 Answers 145 views 0

## Answers ( )

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