Usually the thermal expansion is given as
$$V=V_{0}\alpha \Delta T$$
But the other solution, that is by solving the differential equation of the expansion I get an exponential function. But many source suggest that it is a linear function. Is it linear or exponential.
To be clear, the full form of equation of coefficient is
$$\alpha V = \frac{dV}{dT}$$
And thus solving it we get
$$V=V_{0}e^{\alpha T}$$
Am I wrong? And where I went wrong?
Edit : Sorry about that. V doesn't only signify volume, it signifies length and area too. Sorry for not mentioning it.
I am extremely sorry if I have committed any mistake in this. I am new to this one. Please mention that either.
Best Answer
You have to allow for $\alpha_V$ to have temperature dependence. Take, as an example, the ideal gas. The equation of state implies that $V=Nk_BT/P$ expands linearly with $T$ at constant $P$. Indeed, the volume expansion coefficient is $$ \alpha_V = \frac{1}{V}\frac{dV}{dT}=\frac{1}{T}, $$ inversely proportional to $T$. This means $dV/V=dT/T$, and the expansion is linear.