Question

B. To prevent tank rupture during deep-space travel, an engineering team is studying the effect of temperature on gases confined to small volumes. What is the pressure of 3.00 mol of gas D measured at 251 ∘C in a 1.75-L container assuming ideal behavior?

Express your answer with the appropriate units.

C. There are several real gas equations of state. For this part, real will be approximated by the van der Waals equation of state.

To prevent tank rupture during deep-space travel, an engineering team is studying the effect of temperature on gases confined to small volumes. What is the pressure of 3.00 mol of gas D measured at 251 ∘C in a 1.75-L container assuming real behavior?

Express your answer with the appropriate units.

Answer 1

B. To prevent tank rupture during deep-space travel, an engineering team is studying the effect of temperature on gases confined to small volumes. What is the pressure of 3.00 mol of gas D measured at 251 ∘C in a 1.75-L container assuming ideal behavior?

Express your answer with the appropriate units.

Solution:

We know for Ideal gas, PV = nRT

Therefore, P = nRT/V

                     = (3 mol×8.314 J/K.mol×524 K)/1.75 L

                     = 7468.34 J/L

                     = 7468.34 Kg.m².s^-2/10^-3 m³

                     = 7468.34×10³ Kg.m^-1.s^-2

                     = 7468.34×10³ Pa (Pascal)

C. There are several real gas equations of state. For this part, real will be approximated by the van der Waals equation of state. To prevent tank rupture during deep-space travel, an engineering team is studying the effect of temperature on gases confined to small volumes. What is the pressure of 3.00 mol of gas D measured at 251 ∘C in a 1.75-L container assuming real behavior?

Express your answer with the appropriate units.

Solution:

Van der Waals equation of state for real gas can be given by:

(P + n²a/V²)(V- nb)= nRT -------à(1)

Where, a & b are the parameters are introduced as a measure of average attraction between particles, and the volume excluded from v by a particle respectively.

At high temperature (251 ∘C ) kinetic energy>>attraction energy.

Therefore, n²a/V² can be neglected and equation (1) can be reduced to

P(V- nb)= nRT ; b = roughly the volume of a molecule, (3.5×10^-29 – 1.7 ×10^-28) m³

P = nRT/(V-nb)

   = (3 mol×8.314 J/K.mol×524 K)/1.75 L

   = (3 mol×8.314 J/K.mol×524 K)/(1.75×10^-3 - 3×3.5×10^-29 )m³

   = 13069.6/(1.75×10^-3 - 10.5×10^-29 ) Kg.m^-1.s^-2

   = 13069.6/(1.75 - 10.5×10^-26 ) ×10^-3 Pa

= 13069.6 ×10³/(1.75 - 10.5×10^-26 ) Pa

~ 7468.34×10³ Pa