Applied Physics Tutorial 3 - First Law of Thermodynamics

Before you tackle this tutorial, you might wish to look at Thermal Physics Tutorial 4.

The First Law of Thermodynamics

Thermodynamics is the study of heat flows and how they can be put to work.  Engines work by converting heat energy into movement energy, which can then do useful jobs of work for us.  We need to look at a couple of key words.  A system is the object of interest whose behaviour we are monitoring in relation to its surroundings.  A flask containing gas is a system; the water bath in which the flask is placed is its surroundings.  The diagram below helps to show the idea: The Laws of Thermodynamics were the results of work by nineteenth century physicists.  Ironically the Second Law came before the First Law.  Then a more fundamental law, the Zeroth Law was worked out.

In words the First Law of Thermodynamics is:

The change in internal energy of a system is equal to the sum of energy entering the system through heating and energy entering the system through work done on it.

 Question 1 What is internal energy?

We can write the first law in code: [DQ = heat entering the system; DU = increase in internal energy; DW - work done by the system]

The diagram here explains the idea: Worked Example A lump of lead of mass 0.50 kg is dropped from a height of 20 m onto a hard surface.  It does not bounce but remains at rest. What are DQ, DW, and DU? Answer DQ = 0 J as zero heat is supplied to the system DU = mgDh = 0.5 kg × 9.81 m s-2 × 20 m = 98 J DW = - 98 J as work is done on the system rather than by the system

 Question 2 Some air in a bicycle pump is compressed so that its volume decreases and its internal energy increases.  If 25 J of work are done by the person compressing the air, and if 20 J of thermal energy leave the gas through the walls of the pump, what is the increase in the internal energy of the air? If we compress a gas in a bicycle pump, we find it gets hot.  Then we let the pump cool down, without releasing any gas.

 Question 3 What happens if we release the pump? Consider a cylinder of area A.  A fluid is admitted at a constant pressure, p.  It makes the piston move a distance, s. We know that:

• Pressure (N m-2) = force (N) ÷ area (m2)

p = F/A

• Work done (J) = force (N) × distance moved (m)

W = FDs

Therefore

• Force (N) = pressure (N m-2) × area (m2)

F = pA

• Work done (J) = pressure (N m-2) × area (m2) × distance moved (m)

• Area (m2) × distance moved (m) = change in volume (m3)

So we can write:

Work done (J) = pressure (N m-2) × change in volume (m3)

In code:

DW = pDV

This can be shown in a graph: Question 4 A cylinder has an area of 0.125 m2 .  Steam is admitted at a pressure of 1.5 x 106 Pa.  The piston moves a distance of 0.20 m.  What work is done? Behaviour of Gases

Isothermal Changes

In Unit 5 we saw that the behaviour of ideal gases is governed by the equation:

pV = nRT

If we keep the temperature the same, we can say that pV = constant, which you may remember as Boyle's Law.  Keeping the temperature the same is called an isothermal compression or expansion.  We can sketch a graph: Each of the lines is called an isothermal, because the temperature is kept the same.  We can make the compression and expansion of gases very nearly isothermal by pressing down on a bicycle pump very slowly, so that any heat generated can flow out very slowly.  Similarly we can allow the gas to expand very slowly so that the heat flow in in very slow.

For all isothermal processes:

• pV = constant and p1V1 = p2V2;

• DU = 0 because the internal energy is dependent on temperature, which does not change.

• DQ = DW.  If the gas expands to do work DW, and amount of heat DQ must be supplied

The process is a reversible isothermal change if the piston of the pump is allowed to expand after compression and follows exactly the same line on the graph that it did when being compressed, and ends up in exactly the same place as when it started.

A change where there is no heat flow in or out of a system is called adiabatic.

DQ = 0, therefore DW = DU

If you push the plunger of a bicycle pump in very rapidly and block off the end, you get an adiabatic process where the temperature rise of the gas is entirely due to the work done in compressing the gas.  If a gas is allowed to expand without any heat energy being put in, the process is still adiabatic.  The expansion occurs at the expense of the internal energy.  The gas cools down. An example of this is a little rocket that can be made with a very small cylinder of carbon dioxide at high pressure.  This is shown in the diagram above.  The heat flow through the side is negligible compared with the energy loss that causes the drop in temperature as a result of the expansion of the gas.  The cylinder gets so cold that frost forms on the outside, even though the room is warm.

We can use the gas laws to work out the temperature loss.  A useful equation is: Worked Example A small rocket powered trolley uses a CO2 cartridge which  0.1 mol CO2 gas.  The volume of the cylinder is 15 × 10-6 m3. The temperature of the compressed gas is 300 K.  The gas is compressed to a pressure of 2.2 × 107 Pa.  What is the temperature of the uncompressed gas assuming that 1 mol of gas occupies 24 × 10-3 m3 at 1.01 × 105 Pa?  (R = 8.3 J k-1 mol-1) 0.1 mole occupies 2.4 × 10-3 m3  The equation to use is (2.2 × 107 Pa × 15 × 10-6 m-3) ÷ 300 K = (1.01 × 105 Pa × 2.4 × 10-3 m3) ÷ T2  T2 =  (300 K × 1.01 × 105 Pa × 2.4 × 10-3 m3) ÷ (2.2 × 107 Pa × 15 × 10-6 m3) = 220 K (-53 oC)

 Question 5 Why can we assume that the behaviour of the gas is almost adiabatic?  Temperatures are sometimes given in Celsius.  They must be converted to Kelvin. Watch out for this bear trap.

We can look at the behaviour of a gas being compressed adiabatically.  Remember no heat is allowed to enter or leave the system.  Look at the light green line: The adiabatic line is steeper than the isothermal lines:

• At high pressure low volume, the adiabatic is at the value that you would expect for an isothermal at a high temperature; it has got hot.

• At low pressure, high volume the adiabatic line cuts the isothermal at a low temperature; the gas has become cool.

• The equation for the adiabatic line is: [k - constant; g - ratio Cp/Cv]

• Cv is the energy needed to give unit temperature rise in 1 mole of gas where the volume is kept constant.

• Cp is the energy needed to give unit temperature rise in 1 mole of gas where the pressure is kept constant.

• For a monatomic gas, g = 1.67.

• For a diatomic gas, g = 1.40.

• For a polyatomic gas g = 1.33.

• A more useful version of the equation is: • Since DQ = 0, DW = -DU

 Question 6 The diagram below shows a pump used to inflate a rubber dinghy. When the piston is pushed down, the pressure of air in the cylinder increases until it reaches the pressure of the air in the dinghy.  At this pressure the valve opens and air flows at almost constant pressure into the dinghy. (a) The pump is operated quickly so the compression of the air in the cylinder before the valve opens can be considered adiabatic. At the start of a pump stroke, the pump cylinder contains 4.25 × 10-4 m3 of air at a pressure of 1.01 × 105 Pa and a temperature of 23 °C. The pressure of air in the dinghy is 1.70 × 105 Pa.  Show that, when the valve is about to open, the volume of air in the pump is about 2.9 × 10-4 m3.   g for air = 1.40   (b) Calculate the temperature of the air in the pump when the valve is about to open.   AQA Past Question Isovolumetric Processes

Isovolumetric processes occur at constant volume.  We can show this on a graph that displays the isothermals as we have above. • The process occurs at constant volume.

• p1/T1 = p2/T2

• Since there is no change in volume, no work is done, so all heat entering the gas becomes internal energy.

Isobaric Processes

These happen at a constant pressure.  The graph shows the idea: • The process occurs at constant pressure.

• V1/T1 = V2/T2.

• Some of the heat is used to increase the internal energy, the rest to do work.

 Question 7 The diagram shows a sample of gas enclosed in a cylinder by a frictionless piston of area 150 cm2. When 300 J of energy is supplied to the gas, it expands and does work against a constant pressure of 1.0 x 105 Pa and pushes the piston 16 cm along the cylinder.  Calculate: (a) the work done by the gas (b) the increase in internal energy of the gas.