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Many compressed gases come in large, heavy metal cylinders that are so bevy that they need a special cart to move them around. An 80.0-L tank of nitrogen gas pressurized to 172 arm atm is left in the sun and heats from its normal temperature of 20.0 degrees Celsius to 140.0 degrees Celsius. Determine (a) the final pressure inside the tank and (b) the work, heat, and delta U of the process. assume that behavior is ideal and the heat capacity of diatomic nitrogen is 21.0 j/molk.

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Answer:

a) final pressure inside the tank is 242.4 atm

b)

Work = 0

heat q = 1440.85 kJ

DU = 1440.85 kJ

Explanation:

Given that;

Pressure P1 = 172 atm

Volume V = 80 L

Temperature T1 = 20°C = ( 273.15 + 20) = 293.15 K

Temperature T2 = 140°C = ( 273.15 + 140) = 413.15 K

we know that, gas constant R = 0.0821 atm.L/mol.K

from the Ideal Gas equation;

pV = nRT1

n = pV/RT1

we substitute

n = (172 × 80) / (0.0821 × 293.15)

n = 13760 / 24.067615  

n = 571.72 moles

now

P2 = nRT2/V2

P2 = (571.72 × 0.0821 × 413.15) / 80

P2 = 19392.5222 / 80

P2 = 242.4 atm

Therefore, final pressure inside the tank is 242.4 atm

b)

we know

w = -∫[tex]P_{ext}[/tex] dv

now, since there is no change in volume; dv = 0

so

w = 0

Work = 0

dU = cVDT

Cv = nCr,m

Cv = 571.72 ×  21.0

Cv = 12007.12 J/k

DU = CvΔT

DU = 12007.12 × (413.15 - 293.15)

DU = 1440854.4 J

DU = 1440.85 kJ

DU = q + w

1440.85 = q + 0

heat q = 1440.85 kJ

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