Pressure/Temp/Volume question

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In an isothermal expansion of one mole of an ideal gas from a pressure of 2x10^6 N/m^2 to 2x10^5 N/m^2 at 300K, the change in internal energy (ΔE) is zero due to constant temperature. Consequently, the heat absorbed by the gas (Q) equals the work done (W) during the expansion. The work can be calculated as the area under the PV diagram, but the change in volume is unclear without additional information. To determine the volumes, the ideal gas law (PV=nRT) can be applied to find initial and final volumes based on the given pressures. The entropy change can also be calculated using the formula for reversible processes, but requires the final volume to be known.
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Homework Statement



One mole of an ideal gas at 300K expands isothermally from a pressure of 2x10^6 N/m^2 to a pressure of 2x10^5 N/m^2.

Derive the equations for, and calculate:

the heat absorbed by the gas...
the change in the internal energy of the gas...
the entropy change in the gas if it expands reversibly

Homework Equations



Q=mc(Δt)
ΔE = Q - W
PV = nRT
W = PΔV
Another equation involving ΔE, but i am not sure what it is...

The Attempt at a Solution



Since it is an isothermic equation, there is no Δt...which makes the ΔE equal zero(the internal energy). Since the internal energy is zero, that makes the equation ΔE=Q-W into Q=W. I am not sure how to finish the problem, because i have a change in pressure...but there is no change in temperature...
 
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Work is just the area under a PV diagram. Draw out the diagram, label V1 and V2 (the initial and final volumes), and you'll see what the area is.
 
how do i know what the change in volumes are? all i know is there is a pressure decrease, which could mean anything...unless i am missing something
 

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