Find work done on an ideal gas and a copper block.

[owlman76]

1. Homework Statement

An Ideal gas and a block of copper have equal volumes V and at the same temperature T and atmosphere pressure P. The pressure on both substances is increased reversibly and isothermally to 5P.

a) Find the work done on the ideal gas if V = 0.5 m and T= 300 K P = 1.01*10^5 Pa

b) Find the work done on the copper using the definition of compressibility k= 0.7*10^-6. for copper


2. Homework Equations

PV=nRT
W= PdV
dW=PVBdT - PVkDP


3. The Attempt at a Solution

Part A) P1V1=P2V2=nRT SO W = PdV SO W = nRT/V dV SO W= P1V1 * ln(.1/.5) = -81276.6 Joules

I have no idea if that is correct

Part B) dW=PVBdT - PVkDP

Temp is constant so first part drops out so
dW = -PVkdP

My answer was W=-42844 Joules

Any help on solving these equations is awesome... I don't know if I got either part right or even close to. THANK YOU

 

[Chestermiller]

The answer for the ideal gas is correct, except for the sign which is positive. They are asking for the work done on the gas, not the work done by the gas.

For the copper, $$V=V_0(1-k(P-P_0))$$ so $$dV=-kV_0dP$$and $$dW=-PdV=kV_0PdP$$So, $$W=kV_0(P^2-P_0^2)/2$$ with $$k=7\times 10^{-12}/ Pa^{-1}$$This is because the original k was per bar, not per Pa.

Substituting, we get W = 1.7 J.