91Ó°ÊÓ

Amount of gas $=2.0ml$

Temperature$={30}^{∘}C$

Pressure$=1.5atm$

In an isothermal process, the work is given by

$W=-nRT\ln \left(\frac{V_f}{V_i}\right)$

We know that the final volume is one third of the initial volume; substituting that and the other information we are given, we have

$W=(-2ml×8.314Jâ‹\dots K^{-1}â‹\dots mo{l}^{-1}×{303}^{∘}C\ln \left(\frac{1m^3}{3m^3}\right)=5540\mathrm{J}$

Hence, the amount of work must be done on the gas to compress it to one-third of its initial volume at constant temperature is$5540J.$

Amount of gas$=2.0\mathrm{ml}$

Temperature $=30°\mathrm{C}$

Pressure $=1.5\mathrm{atm}$

(b) In an isobaric process the work is given by

$W=-p\mathrm{Δ}V$

In our case, knowing the relation between the initial and final volumes, we have

$W=-p\left(-\frac{2}{3}V\right)=\frac{2}{3}pV$

We therefore need to find the initial volume from$V:$the state equation, we have

$pV=nRT⇒V=\frac{nRT}{p}$

We can therefore substitute in the expression for the work and find that

$W=\frac{2}{3}p\frac{nRT}{p}=\frac{2}{3}nRT$

In our numerical case, we will have

$W=(\frac{2}{3}×2mol×8.314Jâ‹\dots K^{-1}â‹\dots mo{l}^{-1}×{303}^{∘}C)$

Amount of work must be done on the gas to compress it to one-third of its initial volume at constant pressure is,

$=3360\mathrm{J}$

Amount of gas$=2.0\mathrm{ml}$

Temperature$=30°\mathrm{C}$

Pressure$=1.5\mathrm{atm}$

Below a sketch of such a graph is given:

Therefore the graph of $pV$has been shown.