# 3.3 First law of thermodynamics  (Page 3/6)

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Check Your Understandin g The quantities below represent four different transitions between the same initial and final state. Fill in the blanks.

Q (J) W (J) $\text{Δ}{E}_{\text{int}}\left(\text{J}\right)$
–80 –120
90
40
–40

Line 1, $\text{Δ}{E}_{\text{int}}=40\phantom{\rule{0.2em}{0ex}}\text{J;}$ line 2, $W=50\phantom{\rule{0.2em}{0ex}}\text{J}$ and $\text{Δ}{E}_{\text{int}}=40\phantom{\rule{0.2em}{0ex}}\text{J;}$ line 3, $Q=80\phantom{\rule{0.2em}{0ex}}\text{J}$ and $\text{Δ}{E}_{\text{int}}=40\phantom{\rule{0.2em}{0ex}}\text{J;}$ and line 4, $Q=0$ and $\text{Δ}{E}_{\text{int}}=40\phantom{\rule{0.2em}{0ex}}\text{J}$

## 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 ${p}_{1}=2.10\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}{\phantom{\rule{0.2em}{0ex}}\text{N/m}}^{2}\text{,}\phantom{\rule{0.2em}{0ex}}{p}_{2}=1.05\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}{\phantom{\rule{0.2em}{0ex}}\text{N/m}}^{2}\text{,}\phantom{\rule{0.2em}{0ex}}{V}_{1}=2.25\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}{\phantom{\rule{0.2em}{0ex}}\text{m}}^{3}\text{,}$ and ${V}_{2}=4.50\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}{\phantom{\rule{0.2em}{0ex}}\text{m}}^{3}.$

## 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

${W}_{ABC}={p}_{1}\left({V}_{2}-{V}_{1}\right)=473\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$

Along ADC , the work done by the gas is again the area under the path:

${W}_{ADC}={p}_{2}\left({V}_{2}-{V}_{1}\right)=236\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$

Then using the strategy we just described, we have

${Q}_{ADC}-{Q}_{ABC}={W}_{ADC}-{W}_{ABC},$

${Q}_{ADC}={Q}_{ABC}+{W}_{ADC}-{W}_{ABC}=\left(515+236-473\right)\phantom{\rule{0.2em}{0ex}}\text{J}=278\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$

## Significance

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 $p\text{Δ}V$ . An isothermal line could also have been used, as we have derived the work for an isothermal process as $W=nRT\text{ln}\frac{{V}_{2}}{{V}_{1}}$ .

## Isothermal expansion of an ideal gas

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

1. We saw in the preceding section that the internal energy of an ideal monatomic gas is a function only of temperature. Since $\text{Δ}T=0$ , for this process, $\text{Δ}{E}_{\text{int}}=0.$
2. The quasi-static isothermal expansion of an ideal gas was considered in the preceding section and was found to be
$\begin{array}{cc}\hfill W& =nRT\text{ln}\frac{{V}_{2}}{{V}_{1}}=nRT\text{ln}\frac{3V}{V}\hfill \\ & =\left(1.00\phantom{\rule{0.2em}{0ex}}\text{mol}\right)\left(8.314\phantom{\rule{0.2em}{0ex}}\text{J/K}·\text{mol}\right)\left(300\phantom{\rule{0.2em}{0ex}}\text{K}\right)\left(\text{ln}3\right)=2.74\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}\hfill \end{array}$
3. With the results of parts (a) and (b), we can use the first law to determine the heat added:
$\text{Δ}{E}_{\text{int}}=Q-W=0\text{,}$

$Q=W=2.74\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$

## Significance

An isothermal process has no change in the internal energy. Based on that, the first law of thermodynamics reduces to $Q=W$ .

Check Your Understanding Why was it necessary to state that the process of [link] is quasi-static?

So that the process is represented by the curve $p=nRT\text{/}V$ on the pV plot for the evaluation of work.

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