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10.2The first law of thermodynamics
When an amount of heat Q is given to a gas, the gas will absorb that energy and use it to change its internal energy and/or to do work. Conservation of energy demands that
Q=ÎU+W
where ÎU is the change in internal energy and W is the work done. This formula goes with certain conventions that you must know:
Q > 0 means heat is supplied to the system; Q < 0 means heat is removed
ÎU > 0 means the internal energy and temperature increase; ÎU < 0 means U and T decrease
W > 0 means that the work is done by the gas as it expands; W < 0 means that the work is done on the gas by compressing it.
EXAM TIP
The conventions of signs in the first law are crucial.
EXAM TIP
The first law is just the law of conservation of energy applied to thermodynamics.
CHECK YOURSELF 4
An amount of work equal to 250 J is done on a gas, compressing it, and 250 J of thermal energy is removed from the gas.
Did the temperature of the gas increase, decrease or stay the same?
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WORKED EXAMPLE 10.3
An ideal gas in a container with a piston expands isothermally. Energy Q = 2.0 Ã 105 J is transferred to the gas. Calculate the work done by the gas.
Answer
The process is isothermal so we know that T = constant; it follows that âU = 0 and since
Q=ÎU+W
we must have
W=Q
So, the work done by the gas in this case is equal to the energy supplied to it: 2.0 Ã 105 J.
WORKED EXAMPLE 10.4
An ideal gas expands adiabatically.
aExplain why the temperature will decrease.
bUse your answer in a to explain why the adiabatic curve of an ideal gas expanding from a given state is steeper than the corresponding isothermal curve.
Answer
aUsing the first law Q = âU + W, we have that Q = 0 because the process is adiabatic. Hence
0=ÎU+W
from which we find âU = âW. The gas expands; it is the gas that does the work so W > 0. Therefore âU < 0.
That is, the internal energy and thus temperature decreases. Similarly, if the gas is compressed adiabatically, the temperature will increase.
bThis explains why the adiabatic curve starting from a point on the pressureâvolume diagram is steeper than the isothermal curve starting from that same point (see Figure 10.3). Consider an adiabatic and an isothermal process, both starting from the same point and bringing the gas to the same (expanded) final volume. Since the adiabatic process will reduce the temperature and the isothermal process will not, it follows that the pressure of the final state after the adiabatic expansion will be lower than the pressure of the state reached by the isothermal expansion. Hence, the adiabatic curve must be steeper.
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WORKED EXAMPLE 10.5
An ideal gas with initial volume 0.100 m3 is compressed at constant pressure 3.00 Ã 106 Pa to a volume of 0.080 m3. Find:
athe work done on the gas
bthe heat energy transferred.
Answer
aThe work done is:
W=PÃÎV=3.00Ã106Ã(0.080â0.100)=â6.00Ã104Â J
(The work is negative because the gas is being compressed, work is being done on the gas.)
bFrom the first law:
Q=ÎU+W
So, to find the energy transferred we must first find the change in the internal energy of the gas. Here the pressure is constant so:
ÎU=32PÎV=32(â6.00Ã104)=â9.00Ã104Â J
Finally
Q=â9.00Ã104â6.00Ã104=â1.50Ã105Â J
The negative sign in Q means that this energy was transferred away from the gas.
WORKED EXAMPLE 10.6
Figure 10.8 is a pressureâvolume diagram showing two adiabatics, an isovolumetric and an isobaric process making up a loop ABCDA for an ideal gas.

Figure 10.8: For worked example 10.6.

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aDetermine along which legs energy is supplied (Qin) to the gas or removed (Qout) from the gas.
bFind the relation between Qin, Qout and the net work done in the loop.
Answer
aThere is no energy transferred along the adiabatics BC and DA. Along AB, work is done by the gas so that W > 0 and âU > 0 since the temperature at B is higher than that at A. Hence Q > 0 and energy is supplied to the gas. Along CD, W = 0 (the volume does not change) and âU < 0 because the temperature at D is lower than that at C. Hence Q < 0 and energy is transferred away from the gas.
bApplying the first law for the total change A â A, we see that âU = 0 so
QinÂ +0âQoutÂ +0=0+WnetÂ 
WnetÂ =QinÂ âQoutÂ 
(Wnet is the area of the loop.)
WORKED EXAMPLE 10.7
The loop ABCA in Figure 10.9 consists of an isobaric, an isovolumetric and an isothermal process for a fixed quantity of an ideal gas.

Figure 10.9: For worked example 10.7.

aThe temperature of the gas at A is 320 K. Calculate:
ithe temperature at B
iithe energy transferred from A to B
iiithe energy transferred from B to C.
bThe work done on the gas from C to A is 2.2 kJ. Calculate the net work done in one cycle.
cVerify that the net work done is equal to the heat into the system minus the heat out of the system.
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Answer
ai The temperature at B is found from
VATA=VBTn
that is,
2.0Ã10â3320=8.0Ã10â3TBâTB=1280â1300Â K
iiThe work done by the gas as it expands from A to B is
WAB=PAÎV=8.0Ã105Ã(8.0Ã10â3â2.0Ã10â3)=4800Â J=4.8Â kJ
From the first law, QAB = âUAB + WAB, so we need to find the change in internal energy. Since pressure is constant, ÎU=32PÎV=32Ã4.8Ã103=7.2Ã103Â J. Hence
QAB=4.8+7.2=+12Â kJ.
iiiQBC = âUBC since W = 0 for the change from B to C (the volume is constant). The magnitude of the temperature change from B to C is the same as that from A to B, so the changes in internal energy have the same magnitude (but opposite sign). Therefore QBC = â7.2 kJ.
bThe net work is therefore 4.8 â 2.2 = 2.6 kJ.
cThe heat in is 12 kJ. The heat out is 7.2 kJ along BC and 2.2 kJ along CA.
Therefore QinÂ âQoutÂ =12â(7.2+2.2)=2.6Â kJ=WnetÂ .
TEST YOUR UNDERSTANDING
8The pressure of an ideal gas is halved at constant volume. Was thermal energy added or removed from the gas?
9A quantity of energy Q is supplied to three ideal gases, X, Y and Z. Gas X absorbs Q isothermally, gas Y isovolumetrically and gas Z isobarically. Complete Table 10.2 by inserting the words positive, zero or negative for the work done W, the change in internal energy âU and the temperature change âT of each gas.
W
âU
âT
X
Y
Z
Table 10.2
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10An ideal gas is compressed adiabatically.
aUse the first law of thermodynamics to state and explain the change, if any, in the temperature of the gas.
bExplain your answer to part a by referring to the motion of molecules.
11An ideal gas is kept at constant pressure 6.00 Ã 106 Pa. Its initial temperature is 300 K. The gas expands at constant pressure from a volume of 0.200 m3 to a volume of 0.600 m3. Calculate
athe work done by the gas
bthe temperature of the gas at the new volume
cthe energy taken out of or put into the gas.
12Does the addition of heat Q to a system always result in an increase in the internal energy of the system? Explain.
13In an isothermal expansion of an ideal gas the work done is 3 kJ. Is the heat energy exchanged less than, equal to or greater than 3 kJ?
14A gas expands doing 500 J of work. 500 J of heat are added to the gas. Does the temperature of the gas increase, stay the same or decrease?
15A gas is compressed adiabatically with 500 J of work done on the gas. What is the change in the internal energy of the gas?
16An ideal gas undergoes a change from state P to state Q as shown in Figure 10.10. (The temperature is in kelvin.) For this change, state and explain:
awhether work is done on the gas or by the gas
bwhether energy is supplied to the gas or taken out of the gas.

Figure 10.10: For question 16.

17Two ideal gases, X and Y, have the same pressure, volume and temperature. The same quantity of heat is supplied to both gases. Gas X absorbs the thermal energy at constant volume whereas gas Y absorbs the thermal energy at constant pressure. State and explain which of the two gases will have the highest final temperature.
18Two ideal gases, X and Y, have the same pressure, volume and temperature. Thermal energy is supplied to both gases. Gas X absorbs the thermal energy at constant volume whereas gas Y absorbs the thermal energy at constant pressure. The increase in temperature of both gases is the same. State and explain which of the two gases received the largest quantity of energy.
19An ideal gas changes its state from state A to state B along two different paths I and II as shown in Figure 10.11.

Figure 10.11: For question 19.

In which case is the heat supplied larger?
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20An ideal gas changes its state from state A to state B along two different paths I and II as shown in Figure 10.12.

Figure 10.12: For question 20.

In which case is the thermal energy supplied larger?
21An ideal gas changes its state from state A to state B along two different paths I and II as shown in Figure 10.13.

Figure 10.13: For question 21.

In which case is the thermal energy supplied larger?
22The molar specific heat capacity of an ideal gas is defined as the amount of energy required to change the temperature of one mole of a gas by one kelvin. When n moles of gas absorb energy Q at constant pressure, Q = ncPâT where cP is the molar specific heat capacity at constant pressure.
When n moles of gas absorb energy Q at constant volume, Q = ncVâT where cV is the molar specific heat capacity at constant volume.
Use the first law of thermodynamics to show that cP â cV = R, where R is the universal gas constant.
23A fixed mass of an ideal gas is taken through a cycle ABCA as shown in Figure 10.14.

Figure 10.14: For question 23.

aShow that during the change A â B the energy supplied to the gas is 300 J.
bDuring the change B â C the amount of work done is 100 J and the energy transferred is 490 J. Calculate the change in the internal energy of the gas during the change B â C.
cDeduce the area of the triangle in the diagram.
dHence deduce the pressure at C.
eDetermine the energy transferred during the change C â A.
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24Consider the loop ABCDA shown in Figure 10.15. The temperature at A is 400.

Figure 10.15: for question 24

Calculate:
athe temperature at points B, C and D
bthe change in internal energy along each of the four legs of the cycle
cthe amount of energy transferred to and transferred away from the gas along each leg
dthe ratio of net work done to total thermal energy into the system.
