At `5 xx 10^(5)` bar pressure density of diamond and graphite are `3 g//c c` and `2g // c c` respectively, at certain temperature 'T'. Find the value of `DeltaU - DeltaH` for the conversion of 1 mole of graphite to 1 mole of diamond at temperature 'T' :

At 5xx10^(5) bar pressure density of diamond and graphite are 3 g//c c and 2g//c c respectively, at certain temperature 'T' .Find the value of DeltaU-DeltaH for the conversion of 1 mole of graphite to 1 mole of diamond at temperature 'T' :

At 5xx10^(4) bar pressure density of diamond and graphite are 3 g//c c and 2g//c c respectively, at certain temperature 'T' .Find the value of DeltaU-DeltaH for the conversion of 1 mole of graphite to 1 mole of diamond at temperature 'T' :

At 5xx10^(4) bar pressure density of diamond and graphite are 3 g//c c and 2g//c c respectively, at certain temperature 'T' .Find the value of DeltaU-DeltaH for the conversion of 1 mole of graphite to 1 mole of diamond at temperature 'T' :

At 5xx10^(5) bar pressure, density of diamond and graphite are 3 g//c c and 2g//c c respectively, at certain temperature T. (1 L. atm=100 J)

At 5xx10^(5) bar pressure, density of diamond and graphite are 3 g//c c and 2g//c c respectively, at certain temperature T. (1 L. atm=100 J)

At 500 kbar pressure density of diamond and graphite are 3 "g"//"cc" and 2 "g"//"cc" respectively, at certain temperature T. Find the value of |DeltaH-DeltaU| ("in kJ"//"mole") for the conversion of 1 mole of graphite 1 mole of diamond at 500 kbar pressure. (Given : 1 "bar" = 10^(5) "N"//"m"^(2))

Densities of diamond and graphite are 3 and 2 g mL^(-1) , respectively. The increase of pressure on the equilibrium C_("diamond") hArr C_("graphite")

Densities of diamond and graphite are 3.5 and 2.3 g mL^(-1) , respectively. The increase of pressure on the equilibrium C_("diamond") hArr C_("graphite")

Densities of diamond and graphite are 3.5 and 2.3 g mL^(-1) , respectively. The increase of pressure on the equilibrium C_("diamond") hArr C_("graphite")