Equal volumes of 0.5 mol oxygen and 0.5 mol of nitrogen are mixed at temperature T. The volume of each gas is V . They are to be mixed isothermally to have the volume 2 V. The maximum work done will be

Question

Equal volumes of 0.5 mol oxygen and 0.5 mol of nitrogen are mixed at temperature T. The volume of each gas is V . They are to be mixed isothermally to have the volume 2 V. The maximum work done will be (A) R T log e 2 J (B) R T log e 4 J (C) (1/2) R T log e 4 J (D) (1/2) R T log e 1 J

Tardigrade Admin · Accepted Answer

n=0.5 mol of O 2+0.5 mol of N 2=1 mol In isothermal process, Δ W=n R T ln ((V2/V1))=(1)(R T) ln ((2 V/V))=R T ⋅ log e 2 J

Q. Equal volumes of 0.5 mol oxygen and 0.5 mol of nitrogen are mixed at temperature $T$ . The volume of each gas is $V .$ They are to be mixed isothermally to have the volume $2 V$ . The maximum work done will be

$RT lo g_{e} 2 J$

$RT lo g_{e} 4 J$

$\frac{1}{2} RT lo g_{e} 4 J$

$\frac{1}{2} RT lo g_{e} 1 J$

$n = 0.5 m o l$ of $O_{2} + 0.5 m o l$ of $N_{2} = 1 m o l$
In isothermal process,
$Δ W = n RT ln (\frac{V _{2}}{V _{1}}) = (1) (RT) ln (\frac{2 V}{V}) = RT \cdot lo g_{e} 2 J$

Q. Equal volumes of 0.5 mol oxygen and 0.5 mol of nitrogen are mixed at temperature T. The volume of each gas is V. They are to be mixed isothermally to have the volume 2V. The maximum work done will be

RTloge​2J

RTloge​4J

21​RTloge​4J

21​RTloge​1J