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When a quantity of monatomic ideal gas expands at a constant pressure of 4×104Pa, the volume of the gas increases fromrole="math" localid="1664296449974" 2×10-3m3to 8×10-3m3. What is the change in the internal energy of the gas?

Short Answer

Expert verified

The change in the internal energy would be 360J.

Step by step solution

01

Find a new relation of internal energy

At constant pressure, the formula for the ideal gas is pΔV=nRΔT.

Where nrepresents the number of moles, Ris the gas constant and has the value of 8.314J/mol.K. The change in temperature is due to the change in volume.

The change in internal energy for any gas is represented by

ΔU=nCVΔTΔU=n(32R)ΔT23ΔU=nRΔT

Clearly, we can write 23∆U=p∆U.

02

Put the values and calculate the change in internal energy

Given that p=4×104Pa,V2=8×10-3m3and V1=2×10-3m3.

23ΔU=pΔVΔU=32p(V2-V1)ΔU=32×4×104×(8×10-3-2×10-3)ΔU=360J

So, the change in the internal energy would be 360J.

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