3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally against a constant external pressure of 7.7 kPa until its volume has increased by 2.5 dm'. (b) Calculate the work that would be done if the same expansion occurred reversibly.

(a) In this case, since the initial volume is 18.5 dm³ and the final volume is 21 dm³ (18.5 +2.5), we can compute the work at constant pressure as shown below:

[tex]W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J[/tex]

Which is negative as it expands against the given pressure.

(b) Moreover, of the process is carried out reversibly, the pressure can change, therefore, we need to compute the work via:

[tex]W=nRTln(\frac{V_1}{V_2} )[/tex]

Whereas the moles are computed from the given mass of argon:

[tex]n=6.56g*\frac{1mol}{39.95g}=0.164mol[/tex]

Thus, the work is:

[tex]W=0.164mol*8.314\frac{J}{mol*K} *305Kln(\frac{18.5dm^3}{21dm^3} )\\\\W=-52.8J[/tex]

Regards.

ayfat23 ayfat23 · Answer 2 · 2021-11-04T09:28:14+01:00

The work done at isothermal expansion is [tex]19.25 J[/tex] ,The work done where there is reverse expansion is [tex]52.7J[/tex]

From the question, we were given

as[tex](21 dm^3)[/tex]

But the Initial volume increased by [tex](2.5 dm^3),[/tex]

hence final volume= [tex](21 dm^3) + (2.5 dm^3) = 21dm^3.[/tex]

But[tex]1kpa=1000pa[/tex], and [tex]1cm^3=1000dm^3[/tex]

At constant pressure, we can get the workdone as

W= -(P Δ V)

If substitute the values we have

[tex]=-[(7.7 *1000pa) *(21-18.5)*(0.001m^3)[/tex]

Then, Workdone [tex]= -19.25 J[/tex]

B) incase there is a reverse expansion, the workdone can be gotten as

[tex]W= nRTIn (V1/V2)[/tex]

[tex]R= 8.314J/mole.K[/tex]

[tex]T=305K[/tex]

we can get the moles from mass of argon and its molar mass. The mass= 6.56g, mm= 39.95g

Then number of [tex]mole(n) = (6.56/39.95)* 1 mole= 0.164 mole.[/tex]

Then we can substitute into the formula

[tex]W= - 0.164*8.314*305In(18.5/21)[/tex]

[tex]=-52.7J.[/tex]

Therefore, the work done, when there is reverse expansion is -52.7J

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