Question

a gas obeys the equation of state p(v-b)=RT.for the gas b=0.0391L/mol.calculate the fugacity coefficient for the gas at 1000°c and 1000atm

Answer 1

The fugacity coefficient is
`[(f)/(p) ] = 1.45`

Step-by-step explanation:

From the question we are told that

The gas obeys the equation
`p(v-b) = RT`

The value of b is
`b = b = 0.0391 \ L /mol`

The pressure is
`p = 1000 \ atm`

The temperature is
`T= 1000^oC = 1273 K`

generally

`RT ln[(f)/(p) ] = \int\limits^(p)_(o) [ {v_(r) -v_(i)} ]\, dp`

Where
`(f)/(p)` is the fugacity coefficient

`v_r` is the real volume which is mathematically evaluated from above equation as

`v_r = (RT)/(p) + b`

`v_r = (RT)/(p) + 0.0391`

and
`v_(i)` is the ideal volume which is evaluated from the ideal gas equation (pv = nRT , at n= 1) as

`v_(i) = (RT)/(p)`

So

`RT ln[(f)/(p) ] = \int\limits^(1000)_(o) [[ (RT)/(p) + 0.0391] - [(RT)/(p) ]} ]\, dp`

=>
`RT ln[(f)/(p) ] = \int\limits^(1000)_(o) [0.391 ]\, dp`

=>
`RT ln[(f)/(p) ] = [0.391p]\left | 1000} \atop {0}} \right.`

=>
`RT ln[(f)/(p) ] = 38.1`

So

`ln[(f)/(p) ] = (39.1)/(RT)`

Where R is the gas constant with value
`R = 0.082057\ L \cdot atm \cdot mol^(-1)\cdot K^(-1)`

`[(f)/(p) ] = (39.1)/( 2.303 *0.082057 * 1273)`

`[(f)/(p) ] = 1.45`