Question

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.

Answer 1

Answer:

(a) $W=-19.25J$

(b) $W=-52.8J$

Explanation:

Hello.

(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:

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

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:

$W=nRTln(\frac{V_1}{V_2} )$

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

$n=6.56g*\frac{1mol}{39.95g}=0.164mol$

Thus, the work is:

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

Regards.