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A container with rigid walls is filled with 4.0 mol of air with Cv=2.5R Then the temperature is increased from 17 degrees C to 354 degrees C What is the change in internal energy? Let the ideal-gas constant R = 8.314 J/(mol • K).
337 J
28,000 J
7000 J
2800 J

Respuesta :

MathPhys

Explanation:

Internal energy = heat + work

U = Q + W

Since there's no change in volume (rigid walls), W = 0.

U = Q

U = n Cᵥ ΔT

U = (4.0 mol) (2.5 × 8.314 J/mol/K) (354 C − 17 C)

U = 28,000 J

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