Recent content by A Story of a Student

I found they to be same. ##\log(\frac{1}{z})=\log(-i)=\ln(1)+i\arg(-i)=-i\frac{\pi}{2}## ##-\log(z)=-\ln(1)-i\arg(i)=-i\frac{\pi}{2}##

I have reached a conclusion that no such z can be found. Are there any flaws in my argument? Or are there cases that aren't covered in this? Attempt ##\log(\frac{1}{z})=\ln\frac{1}{|z|}+i\arg(\frac{1}{z})## ##-\log(z)=-\ln|z|-i\arg(z)## For the real part...

If the integration of post 9 were right, does it imply that for $$\Delta T C_{V,ideal}=\Delta U$$ since dU/dV=0; for vdw gas, we have an extra positive term added on the dU side that makes it bigger? $$\Delta T C_{V,vdw}=\Delta U+\frac{n^2a}{V^2}$$

Sorry I did not follow. For $$-[P-T\left(\frac{\partial P}{\partial T}\right)_V]=-[\frac{nRT}{V-nb}-\frac{n^2a}{V^2}-T\frac{nR}{V-nb}]=\frac{n^2a}{V^2}$$ right? Even dropping the n will give positive results. So taking the partial derivatives with respect to T means Cv for vdw gas does not...

I have tried to do that, I got $$dU=C_VdT+\frac{n^2a}{V^2}dV$$ and the integration (which I am not sure if this is correct) gives me $$\Delta U=C_V\Delta T-\frac{n^2a}{V}$$ Is so far the calculation correct?

Yes! We have derived that as a homework assignment. However, how do I go from this to heat capacity? Btw thanks for the help!

$$dU=C_V dT+\left(\frac{\partial U}{\partial V}\right)_{T}dV$$

I think the C_V for van der waals gas will be larger than ideal gas since it‘s a more precise description. However, for the relationship I cannot come up with a specific equation.