The densities of graphite and diamond at `298K` are `2.25` and `3.31gcm^(-3)` , respectively. If the standard free energy difference `(DeltaG^(0))` is equal to `1895Jmol^(-1)` , the pressure at which graphite will be transformed into diamond at `298K` isA. `9.92xx106Pa`B. `9.92xx105Pa`C. `9.92xx108Pa`D. `9.92xx107Pa`

The densities of graphite and diamond at `298K` are `2.25` and `3.31gc

1.	The densities of graphite and diamond at `298K` are `2.25` and `3.31gcm^(-3)` , respectively. If the standard free energy difference `(DeltaG^(0))` is equal to `1895Jmol^(-1)` , the pressure at which graphite will be transformed into diamond at `298K` isA. `9.92xx106Pa`B. `9.92xx105Pa`C. `9.92xx108Pa`D. `9.92xx107Pa`
Answer» Correct Answer - B Volume of graphite `=("mass")/("density")=(112)/(2.25)` Volume of diamond `=(12)/(3.31)` Change in volume, `DeltaV=((12)/(3.31)-(12)/(2.25))xx10^(-3)L=-1.91xx10^(-3)L` `DeltaG^(@)` =work done `=-pDeltaV` `P=(DeltaG^(@))/(DeltaV)` `=(1895Jmol^(-1))/(1.91xx10^(_3)xx 101.3)=9794` `" " [because 1 "atm" =10^(5) xx 1.013 "pa"]` `=9.92 xx 10^(8)` Pa

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