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Simora [160]
3 years ago
14

A gas contained within a piston-cylinder assembly, initially at a volume of 0.1 m3 , undergoes a constant-pressure expansion at

2 bars to a final volume of 0.12 m3 , while being slowly heated through the base. The change in internal change of the gas 0.25 kJ. The piston and cylinder walls are fabricated from heat-resistant material and the piston moves smoothly in the cylinder. For the gas as the system, evaluate work and heat transfer each in kJ (Neglect the potential energy change and kinetic energy change).
Physics
1 answer:
Gnom [1K]3 years ago
7 0

Answer:

Work: 4.0 kJ, heat: 4.25 kJ

Explanation:

For a gas transformation at constant pressure, the work done by the gas is given by

W=p(V_f -V_i)

where in this case we have:

p = 2 bar = 2\cdot 10^5 Pa is the pressure

V_i = 0.1 m^3 is the initial volume

V_f = 0.12 m^3 is the final volume

Substituting,

W=(2\cdot 10^5)(0.12-0.10)=4000 J = 4.0 kJ

The 1st law of thermodynamics also states that

\Delta U = Q-W

where

\Delta U is the change in internal energy of the gas

Q is the heat absorbed by the gas

Here we know that

\Delta U = +0.25 kJ

Therefore we can re-arrange the equation to find the heat absorbed by the gas:

Q=\Delta U + W = 0.25 kJ + 4.0 kJ = 4.25 kJ

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