91Ó°ÊÓ

A cylinder has oxygen =0.250 mol

Pressure=2.40 x 10⁵Pa

Temperature=355K

Take oxygen as ideal gas. Which is first expands isobarically to twice its original volume. It is then compressed isothermally back to its original volume, and finally it is cooled isochorically to its original pressure.

(a)The pressure versus volume or PV- diagram is shown,

Here process from a →b is isobaric expansion. In process from b→c is isothermal compression and from c→a is isochoric cooling. The initial value of volume is V0. at the end of isobaric expansion, the volume is 2V0. The value of maximum pressure for ideal gas is given as,

pV = nRT

(b)the temperature during the isothermal compression.

As,

pV = nRT

from a→b, the value of p and n do not change.

$\frac{\mathrm{V}}{\mathrm{T}}=\frac{\mathrm{nR}}{\mathrm{p}}=\mathrm{constant}$

take V1=V0 and T1=355 K are the initial volume and temperature respectively also V2=2V0 is the final volume and T2 is the final temperature. So,

$\frac{{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{v}}_2}{{\mathrm{T}}_2}→{\mathrm{T}}_2={\mathrm{T}}_1\left(\frac{{\mathrm{v}}_2}{{\mathrm{v}}_1}\right)$

Put all the values,

${\mathrm{T}}_2=355×\frac{2{\mathrm{V}}_0}{{\mathrm{V}}_0}=710\mathrm{K}$

Thus, the temperature during the isothermal compression is 710 K

(c)the maximum pressure.

temperature T is constant in isothermal compression. take,

$$\begin{gathered} {\mathrm{p}}_1{\mathrm{V}}_1={\mathrm{p}}_2{\mathrm{V}}_2 \\ {\mathrm{p}}_2={\mathrm{p}}_1\left(\frac{{\mathrm{V}}_1}{{\mathrm{V}}_1}\right) \end{gathered}$$

in isothermal process,

$$\begin{gathered} {\mathrm{V}}_1=2{\mathrm{V}}_0 \\ {\mathrm{V}}_2={\mathrm{V}}_0 \end{gathered}$$

calculate maximum value of the pressure by,

${\mathrm{p}}_2=2.40×{10}^5\mathrm{pa}\left(\frac{2{\mathrm{V}}_0}{{\mathrm{V}}_0}\right)=4.80×{10}^5\mathrm{pa}$

Thus,the maximum pressure is4.80×10⁵ Pa

(d) the total work done by the piston on the gas during the series of processes.

Work done is given by,

${\mathrm{W}}_1=\mathrm{p}∆\mathrm{V}=-\mathrm{nR}∆\mathrm{T}$

Put all the given values,

${\mathrm{W}}_1=-\left(0.250\right)×\left(710-355\right)=-737.5\mathrm{J}$

Work done during the isothermal compression is,

$$\begin{gathered} {\mathrm{W}}_2=-\mathrm{nR}∆{\mathrm{Tlog}}_{\mathrm{e}}\mathrm{e}\left(\frac{{\mathrm{V}}_2}{2{\mathrm{V}}_1}\right) \\ {\mathrm{W}}_2=-0.250×\left(8.31\right)×\left(710\right){\log }_{\mathrm{e}}\mathrm{e}\left(\frac{{\mathrm{V}}_0}{2\mathrm{V}0}\right) \\ {\mathrm{W}}_2=1022.4\mathrm{J} \end{gathered}$$

Work done during the isochoric cooling process is zero because there is no change in volume. So, the total work done by the piston on the gas is given as,

$$\begin{gathered} \mathrm{W}={\mathrm{W}}_1+{\mathrm{W}}_2+{\mathrm{W}}_3 \\ \mathrm{W}=\left(-737.5+1022.4+0\right)\mathrm{J} \\ \mathrm{W}=284.9\mathrm{J} \end{gathered}$$

Thus, the total work done by the piston on the gas during the series of processes is 284.9 J