Need help with finding the work done on an ideal gas and copper block

[owlman76]

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

Homework Equations

PV=nRT
W= PdV
dW=PVBdT - PVkDP

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

 

[mark726]

Thank you for your post. I would like to provide some guidance on solving these equations and finding the correct solutions for both parts A and B.

Part A:

To find the work done on the ideal gas, we can use the equation W = PdV. However, we need to first find the change in volume (dV) in order to solve for the work done. We can do this by rearranging the ideal gas law, PV=nRT, to solve for V:

V = nRT/P

Now, we can plug in the given values for V, T, and P to find the initial volume (V1) of the gas:

V1 = (1 mol * 8.314 J/mol*K * 300 K) / (1.01*10^5 Pa) = 2.47*10^-2 m^3

The final volume (V2) can be found using the same equation, but with the new pressure (5P):

V2 = (1 mol * 8.314 J/mol*K * 300 K) / (5 * 1.01*10^5 Pa) = 4.94*10^-3 m^3

Now, we can plug these values into the equation W = PdV to find the work done:

W = (5 * 1.01*10^5 Pa) * (4.94*10^-3 m^3 - 2.47*10^-2 m^3) = -81276.6 Joules

Therefore, your answer for part A was correct!

Part B:

To find the work done on the copper, we can use the equation dW = -PVkdP. We need to solve for dW, so we can integrate this equation to find the total work done:

W = ∫ -PVkdP

To integrate this equation, we need to know the initial and final pressures (P1 and P2) and the compressibility (k) of copper. We can use the given value of k= 0.7*10^-6 to solve for dW:

dW = -∫ P * (0.7*10^-6) * dP

Integrating this equation from P1= 1.01*10^5 Pa to P2= 5 * 1.01*10^5