Sketch shows the plot of Z vs p for a hypothetical gas for one mole at three distinct temperature Boyle's temperature is the temperature at which a gas shows ideal behaviour over a pressure range in the low pressure region. Boyle's temperature (Tb)=(a/R b). If a plot is obtained at temperature below Boyle's temperature then the curve will show negative deviation in low pressure region and positive deviation in the high pressure region. Near critical temperature, the curve is more likely as CO 2 and the temperature above critical temperature curve is more like H 2 at 0° C For 500 K plot value of Z changes from 2 to 2.2 if pressure is varied from 1000 atm to 1200 atm (high pressure) then the value of (b/R T) will be

Question

Sketch shows the plot of Z vs p for a hypothetical gas for one mole at three distinct temperature <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/75d34ff4e49db1cb63bf0dabc96370a8-.png" /> Boyle's temperature is the temperature at which a gas shows ideal behaviour over a pressure range in the low pressure region. Boyle's temperature (Tb)=(a/R b). If a plot is obtained at temperature below Boyle's temperature then the curve will show negative deviation in low pressure region and positive deviation in the high pressure region. Near critical temperature, the curve is more likely as CO 2 and the temperature above critical temperature curve is more like H 2 at 0° C For 500 K plot value of Z changes from 2 to 2.2 if pressure is varied from 1000 atm to 1200 atm (high pressure) then the value of (b/R T) will be (A) 10-4 atm -1 (B) 10-3 atm -1 (C) 10-5 atm -1 (D) 0.10 atm -1

Tardigrade Admin · Accepted Answer

We know that In the high pressure region Z=1+(p b/R T) 2=1+(1000 b/R T) ldots text (i) 2.2=1+(1200 b/R T) ldots text ..ii) Solving both the equation we get (b/R T)=10-3 atm -1

$1 0^{- 4} a t m^{- 1}$

$1 0^{- 3} a t m^{- 1}$

$1 0^{- 5} a t m^{- 1}$

$0.10 a t m^{- 1}$

We know that
In the high pressure region
$Z = 1 + \frac{p b}{RT}$
$2 = 1 + \frac{1000 b}{RT} \dots (i)$
$2.2 = 1 + \frac{1200 b}{RT} \dots ..ii)$
Solving both the equation we get $\frac{b}{RT} = 1 0^{- 3} a t m^{- 1}$

10−4atm−1

10−3atm−1

10−5atm−1

0.10atm−1

We know that In the high pressure region Z=1+RTpb​ 2=1+RT1000b​… (i) 2.2=1+RT1200b​…..ii) Solving both the equation we get RTb​=10−3atm−1

$1 0^{- 4} a t m^{- 1}$

$1 0^{- 3} a t m^{- 1}$

$1 0^{- 5} a t m^{- 1}$

$0.10 a t m^{- 1}$

We know that
In the high pressure region
$Z = 1 + \frac{p b}{RT}$
$2 = 1 + \frac{1000 b}{RT} \dots (i)$
$2.2 = 1 + \frac{1200 b}{RT} \dots ..ii)$
Solving both the equation we get $\frac{b}{RT} = 1 0^{- 3} a t m^{- 1}$