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Tutorial 1 Notes

# Tutorial 1 Notes - CHM2132 tutorial notes for Sept 11th...

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CHM2132 tutorial notes for Sept. 11 th , 15 th . 1. What is the slope of the curve f ( x ) = 3ln x + 2 x ! 4 x 3 at the point x =3? Since the question is asking for the slope of a curve at a specific point we need to know the slope of the tangent to the curve at that point – the derivative of f ( x , y , z ) . d f ( x ) [ ] dx = d 3ln x + 2 x ! 4 x 3 # % dx Some handy general formulas for derivatives that we will need to solve this are: dx n dx = nx n ! 1 and d ln x dx = 1 x (These are also listed in the inside cover of the text.) Now the derivative is: d f ( x ) [ ] dx = 3 x ! 2 x 2 ! 12 x 2 Substitute in x =3: d f ( x ) [ ] dx = 3 3 ! 2 3 2 ! 12 3 ( ) 2 = ! 107 Therefore the slope of the curve is -107 at x =3. 2. Determine the expression for an ideal gas that describes how the temperature of a gas changes with the pressure for a fixed amount of gas at a constant volume. We know how the temperature of an ideal gas is related to its other variables by the ideal gas equation: T = pV nR The question is, what happens when we change the pressure and keep everything else constant? Mathematically this corresponds to a derivative – we need to know the derivative of T with respect to p . However, there are more variables that define this

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Tutorial 1 Notes - CHM2132 tutorial notes for Sept 11th...

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