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A gas is compressed at a constant pressure of 0.643 atm from 5.46 L to 2.01 L. In the process, 564 J of energy leaves the gas by heat. What is the work done on the gas? Remember: 1 atm = 1.013 × 105 Pa. Answer in units of J. 008 (part 2 of 2) 10.0 points What is the change in its internal energy? Answer in units of J.

Respuesta :

Answer:

Part 1)

- 224.6 J

Part 2)

- 339.4 J

Explanation:

[tex]P[/tex] = Constant pressure acting on gas = 0.643 atm = (0.643) (1.013 x 10⁵) Pa = 0.651 x 10⁵ Pa

[tex]V_{i}[/tex] = initial volume = 5.46 L = 0.00546 m³

[tex]V_{f}[/tex] = final volume = 2.01 L = 0.00201 m³

Work done on the gas is given as

[tex]W = P (V_{f} - V_{i})\\W = (0.651\times10^{5}) ((0.00201) - (0.00546))\\W = - 224.6 J[/tex]

Part 2)

[tex]\Delta U[/tex] = Change in the internal energy

[tex]Q[/tex] = Heat energy escaped = - 564 J

Using First law of thermodynamics

[tex]Q = W + \Delta U\\- 564 = - 224.6 +  \Delta U\\ \Delta U = - 339.4 J[/tex]

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