Check Your Understanding The quantities below represent four different transitions between the same initial and final state. Fill in the blanks.
Q (J)
W (J)
$\text{\Delta}{E}_{\text{int}}(\text{J})$
–80
–120
90
40
–40
Line 1,
$\text{\Delta}{E}_{\text{int}}=40\phantom{\rule{0.2em}{0ex}}\text{J;}$ line 2,
$W=50\phantom{\rule{0.2em}{0ex}}\text{J}$ and
$\text{\Delta}{E}_{\text{int}}=40\phantom{\rule{0.2em}{0ex}}\text{J;}$ line 3,
$Q=80\phantom{\rule{0.2em}{0ex}}\text{J}$ and
$\text{\Delta}{E}_{\text{int}}=40\phantom{\rule{0.2em}{0ex}}\text{J;}$ and line 4,
$Q=0$ and
$\text{\Delta}{E}_{\text{int}}=40\phantom{\rule{0.2em}{0ex}}\text{J}$
An ideal gas making transitions between two states
Consider the quasi-static expansions of an ideal gas between the equilibrium states
A and
C of
[link] . If 515 J of heat are added to the gas as it traverses the path
ABC , how much heat is required for the transition along
ADC ? Assume that
${p}_{1}=2.10\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}{\phantom{\rule{0.2em}{0ex}}\text{N/m}}^{2}\text{,}\phantom{\rule{0.2em}{0ex}}{p}_{2}=1.05\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}{\phantom{\rule{0.2em}{0ex}}\text{N/m}}^{2}\text{,}\phantom{\rule{0.2em}{0ex}}{V}_{1}=2.25\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-3}}{\phantom{\rule{0.2em}{0ex}}\text{m}}^{3}\text{,}$ and
${V}_{2}=4.50\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-3}}{\phantom{\rule{0.2em}{0ex}}\text{m}}^{3}.$
Strategy
The difference in work done between process
ABC and process
ADC is the area enclosed by
ABCD . Because the change of the internal energy (a function of state) is the same for both processes, the difference in work is thus the same as the difference in heat transferred to the system.
Solution
For path
ABC , the heat added is
${Q}_{ABC}=515\phantom{\rule{0.2em}{0ex}}\text{J}$ and the work done by the gas is the area under the path on the
pV diagram, which is
The work calculations in this problem are made simple since no work is done along
AD and
BC and along
AB and
DC ; the pressure is constant over the volume change, so the work done is simply
$p\text{\Delta}V$ . An isothermal line could also have been used, as we have derived the work for an isothermal process as
$W=nRT\text{ln}\frac{{V}_{2}}{{V}_{1}}$ .
Heat is added to 1 mol of an ideal monatomic gas confined to a cylinder with a movable piston at one end. The gas expands quasi-statically at a constant temperature of 300 K until its volume increases from
V to 3
V . (a) What is the change in internal energy of the gas? (b) How much work does the gas do? (c) How much heat is added to the gas?
Strategy
(a) Because the system is an ideal gas, the internal energy only changes when the temperature changes. (b) The heat added to the system is therefore purely used to do work that has been calculated in
Work, Heat, and Internal Energy . (c) Lastly, the first law of thermodynamics can be used to calculate the heat added to the gas.
Solution
We saw in the preceding section that the internal energy of an ideal monatomic gas is a function only of temperature. Since
$\text{\Delta}T=0$ , for this process,
$\text{\Delta}{E}_{\text{int}}=0.$
The quasi-static isothermal expansion of an ideal gas was considered in the preceding section and was found to be
Radio Stations often advertis "instant news,,if that meens you can hear the news the instant the radio announcer speaks it is the claim true? what approximate time interval is required for a message to travel from Cairo to Aswan by radio waves (500km) (Assume the waves Casbe detected at this range )