\(\frac{2}{T}\)
\(\frac{3}{T}\)
\(\frac{1}{2T}\)
\(\frac{1}{T}\)
Given:
\[ \frac{dV}{V} = \alpha \frac{dT}{T} \]
For a constant \( P T^2 \), we have:
\[ \frac{dV}{dT} = \left( C \right) \frac{3}{T^2} \]
So, the volume expansion coefficient is:
\[ \frac{dV}{V} = \frac{3}{T^2} \cdot dT \]
List - I | List -II | ||
---|---|---|---|
a | Isothermal | i | Pressure constant |
b | Isobaric | ii | Temperature constant |
c | Adiabatic | iii | Volume constant |
d | Isobaric | iv | Heat content is constant |
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32