# Problems In Highschool Chemistry/PhChem/thermo1/advsiso

## Contents

## Question 1[edit]

**Which of the two – adiabatic or isothermal – processes has a steeper PV graph?**

### Instructions[edit]

If we are talking about the nature of a graph, we need to consider the mathematical nature of it too. So, an adiabatic process is represented as PV^{y} = constant on a PV graph, while an isothermal process is represented by PV = constant.

Carefully note that the graph of the isothermal process is a rectangular hyperbola on a PV graph, while that of an adiabatic process *resembles* a hyperbola.

Now, the question is, which Process takes a greater dip down in the PV curve? Visualize what is being asked. Take the question literally. If we are talking about an expansion, which process will register a greater dip in pressure for the same amount of increase in volume? The answer can be obtained by merely looking at the equations.

### Arguments[edit]

Suppose the volume increases by some factor, say m times. In an adiabatic process, the pressure then HAS TO dip by a factor of m^{y} to keep the LHS constant. Do the math for an isothermal process, for an increase in volume by a factor of m times, the pressure dips down by the same factor. This means, that the pressure registers a greater dip in adiabatic process than in an isothermal process for the same volume. Note that this is irrespective of the value of the constant on the RHS.

Try to grasp what has been said, and draw an accompanying graph to this point in your notebook.

### Important Note[edit]

Remember that m^{y} is greater than m only if y>1. This is always true, because the adiabatic exponent of any gas is always greater than unity. But suppose, if we were dealing with a process like PV^{1/2} = constant, its graph would be less steep than the isothermal process.

Are you confused? Please do not be. Read the note below, and move ahead only after you are sure that you got the essence of what has been said.

### Concluding Notes[edit]

This is not an uncommon logic, and a common sense dealing with them should evolve in you during an introductory course in the sciences. The whole logic explained above is nothing but one line, what LHS would tend to raise more?

Now suppose, if you are given adiabatic curves of two ideal gases, one monoatomic and the other diatomic, initiating through the same state and registering the same increase in volume. How would you tell which graph belongs to which gas? In other words, what gas would end up with a lower pressure and what would end up with a higher pressure and why?

The answer, of course, is the monoatomic gas will have the lower pressure. But it is up to you to convince yourself about this using the same logic discussed above.

Hint : The value of gamma for the monoatomic gas is greater than that for a diatomic gas.