The pressure-volume of varies thermodynamic process is shown in graphs:

Work is the mole of transference of energy. It has been observed that reversible work done by the system is the maximum obtainable work.
`w_(rev) gt w_(irr)`
The works of isothermal and adiabatic processes are different from each other.
`w_("isothermal reversible") = - 2.303 nRT log_(10) ((V_(2))/(V_(1)))`
`= 2.303 nRT log_(10)((P_(2))/(P_(1)))`
`w_("adiabatic reversible") = C_(V) (T_(1)-T_(2))`
Calculate work done when `1` mole of an ideal gas is expanded reversibly from `30L` to `60L` at a constant temperature of `300K`

Explanation

Molar conductivity is given by the relation:
`wedge_(m) =(1000 k)/(M)`
Let us calculate the molar conductivity at all five concentrations given above:
`wedge_(m_(1)) (m_(1)=0.001M) =(k(Sm^(-1)))/("molarity"("mol L"^(-1)) xx 1000"L m"^(-3))`
`=(1.237 xx 10^(-2))/( 0.001 xx 1000) Sm^(2) "mol"^(-1)`
`=1.237 xx 10^(-2) Sm^(-2) "mol"^(-1) =123.7S cm^(2)"mol"^(-1)`
`wedge_(m_(2)) (m_(2)=0.01M) =(11.85 xx 10^(-2))/(0.01 xx 1000) Sm^(2) "mol"^(-1)`
`=11.85 xx 10^(-3) Sm^(2) mol^(-1) = 118.5 S cm^(2) "mol"^(-1)`
`wedge_(m_(3)) (m_(3)=0.02M) =(23.15xx10^(-2))/(0.02 xx 1000)`
`=11.57 xx 10^(-3)Sm^(2) "mol"^(-1) = 115.7"S cm"^(2)"mol"^(-1)`
`wedge_(m_(4)) (m_(4)=0.05M)= (55.53 xx 10^(-2))/(0.05 xx 1000) `
`=11.10 xx 10^(-3) Sm^(2) "mol"^(-1) =111.0"S cm"^(2)"mol"^(-1)`
`wedge_(m_(5)) (m_(5)= 0.10M)=(106.74xx10^(-2))/(0.1 xx 1000)`
`=10.67 xx 10^(-3)Sm^(2)"mol"^(-1) =106.7Scm^(2)"mol"^(-1)`
On tabulating these values against `sqrt(c )`, we get

On plotting `wedge` against `c^(1//2)`, a graph shown in Fig. will be obtained. On extrapolation, `wedge_(m)` is nearly 130 S `cm^(2)"mol"^(-1)`.