A monoatomic ideal gas undergoes an isothermal expansion at 300 K, as the volume increased from 0.010 m^3 to 0.040 m^3. The final pressure is 130 kPa. What is t...
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Answer:
To determine the change in the internal energy of the system, we use the first law of thermodynamics which expresses the change in internal energy. It is expressed as:
ΔU = Q + W
where ΔU is the change in total energy
Q is the heat energy
W is work
It is said that the system undergoes an isothermal process so the value of Q would be zero. The change in internal energy would be equal to work. In thermodynamics, work is expressed as:
Work = ∫ - PdV
Since P is not constant, we need to express it in terms of V and substitute it to the equation. We use PV = nRT
P = nRT/V
Work = ∫ - (nRT/V)dV
Integrating from V1 to V2, we will have:
Work = nRT ln V2/V1
ΔU = nRT ln (V2/V1) = 1 (8.314) (300) ln (0.040 / 0.010)
ΔU = 34457.40 J
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