To avoid needlessly long descriptions, we can use diagrams to show the processes that take place in a container of gas.
Key Concepts
A graph of pressure vs volume (\(pV\) diagram) enables us to show all possible types of process because of the interrelationship between pressure, volume and temperature in the ideal gas law: \(pV\propto T\)
We can use the first law to consider the changing heat, internal energy and work done: \(Q=\Delta U+p\Delta V\)
An isovolumetric process takes place at constant volume. This is represented by a vertical line (isochore) on a \(pV\) diagram.
In this example, \(\textrm{B}\to\textrm{C}\) represents an increase in pressure and temperature.
\(p\propto T\)
- \(p\) is pressure (Pa)
- \(T\) is temperature (K)
\({p_1\over T_1}={p_2\over T_2}\)
Since internal energy is increasing, heat must be gained by the gas. No work is done.
An isobaric process takes place at constant pressure. This is represented by a horizontal line (isobar) on a \(pV\)diagram.
In this example, \(\textrm{A}\to\textrm{B}\) represents an expansion and increase in temperature.
\(V\propto T\)
- \(V\) is volume (m3)
- \(T\) is temperature (K)
\({V_1\over T_1}={V_2\over T_2}\)
Since internal energy is increasing and work is done on the surroundings, heat must be gained by the gas.
A
n isothermal process takes place at constant temperature. This is represented by a line of inverse proportion (isotherm) on a \(pV\) diagram.
In this example, \(\textrm{E}\to\textrm{F}\) represents an expansion with a reduction in pressure.
\(p\propto {1\over V}\)
- \(p\) is pressure (Pa)
- \(V\) is volume (m3)
\(p_1V_1=p_2V_2\)
Since internal energy is constant and work is done on the surroundings, heat must be gained by the gas. Isothermal processes are conducted slowly, to avoid increasing the kinetic energy of the particles.
No heat is gained or lost during an adiabatic process. This is represented by an adiabat on a \(pV\) diagram, which has a steeper gradient throughout than an isotherm.
In this example, \(\textrm{G}\to\textrm{H}\) represents an expansion with a reduction in pressure and temperature.
The following equation can be derived for monatomic gases:
\(p\propto {1\over V^{5\over 3}}\)
- \(p\) is pressure (Pa)
- \(V\) is volume (m3)
\(p_1V_1^{5\over 3}=p_2V_2^{5\over 3}\)
Since no heat enters or leaves, the decrease in internal energy must be equal to the work done on the surroundings. Adiabatic processes are conducted rapidly, to prevent sufficient time for heat to enter or leave.
Use flashcards to practise your recall.
Use quizzes to practise application of theory.
START QUIZ!
This \(pV\) graph shows two isochoric changes.
The ratio \(Q_{AB} \over Q_{CD}\) is
Since volume is constant, \(Q = ΔU = {3\over2}nRΔT\Rightarrow Q\propto \Delta T\)
\(ΔT\) is the same for both as the pressure triples in both cases (and \(\Delta T\propto \Delta p\)).
This \(pV\) graph shows two isochoric changes. The amount of gas is such that \(nR =10\) kPa cm3 K-1.
The change in temperature from A to B is:
Recall the ideal gas law: \(PV=nRT\)
This needs to be used separately at each point:
- \(T_A = (100 \times 100)\div 10\)
- \(T_B = (100 \times 300)\div 10\)
This \(pV\) graph shows two isobaric changes.
The ratio \(W_{AB}\over W_{CD}\) is:
Work done is equal to the area under the line.
\(W_{AB} = 200\times10^3 \times 150\times10^{-6} = 30\text{ J}\)
\(W_{CD} = 100\times10^3 \times 300\times10^{-6} = 30\text{ J}\)
NOTE: Including the unit prefixes was not required here as the question required a ratio.
This \(pV\) graph shows two isobaric changes.
The work done by the gas expanding from C to D is:
Work done is equal to the area under the line.
\(W_{CD} = 100\times10^3 \times 300\times10^{-6} = 30\text{ J}\)
NOTE: Take care with unit conversions involving volume.
This \(pV\) graph shows two isobaric changes. The amount of gas is such that \(nR =0.01\) Pa m3 K-1.
The temperature change in each case is 3000K. How much heat is added expanding from C to D?
Internal energy: \(ΔU = {3\over2} nRΔT = 45 \text{ J}\)
Work done by the gas in expansion: \(W_{CD} = 100\times10^3 \times 300\times10^{-6} = 30\text{ J}\)
Using the first law of thermodynamics: \(Q = ΔU + W\)
This \(pV\) graph represents two isothermal changes.
The ratio \(W_{AB}\over W_{CD}\) is approximately:
Work done is equal to the area under the graph. In this case we need to count squares.
\(A\rightarrow B\) 6.5 units
\(C\rightarrow D \) 1.5 units
\(6.5\div1.5\approx4\)
NOTE: If required to calculate exact values for work done, you would need to calculate the work represented per square.
This \(pV\) graph represents two isothermal changes.
The ratio \(Q_{AB}\over Q_{CD}\) is approximately:
Since temperature is constant (isothermal), there is no change in internal energy. Using the first law, \(Q=W\)
Work done is equal to the area under the graph. In this case we need to count squares.
\(A\rightarrow B\) 6.5 units
\(C\rightarrow D \) 1.5 units
\(6.5\div1.5\approx4\)
This \(pV\) graph represents two isothermal changes.
The ratio \(T_B\over T_C\) is:
Note that \(T_B=T_A\) as these changes occur on isotherms.
Using the ideal gas law, \(pV=nRT\):
\({T_A\over T_C}={P_A\over P_C}={400\over 100}\)
HINT: It can be helpful to write \(pVT\) values in pencil at each location on the graph as this will make it easier to spot the equations you can use.
This \(pV\) graph shows an adiabatic expansion. The amount of gas is such that \(nR =0.01\) Pa m3 K-1.
The work done from A to B is 24 J. What is the temperature at B?
An adiabatic change means \(Q = 0 \Rightarrow ΔU = - W\):
\({3\over 2}nRΔT = -24\)
\(ΔT = -1600 \text{ K}\Rightarrow T_B=3000-1600=1400\text{ K}\)
This \(pV\) graph shows an adiabatic expansion. The amount of gas is such that \(nR =0.01\) Pa m3 K-1.
The work done from A to B is 24 J. How much heat is added?
An adiabatic process means that no heat is exchanged (in or out).
This \(pV\) graph shows an adiabatic expansion.
The ratio \(p_AV_A^{5\over 3}\over p_BV_B^{5\over 3}\) is:
\(pV^{5\over 3}\) is constant for adiabatic processes.
HINT: This must be used instead of \({p_AV_A\over T_A}={p_BV_B\over T_B}\) when processes are adiabatic.
How much of Processes have you understood?
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