What is the change in enthalpy for this process? You are given a closed vessel a

Question

What is the change in enthalpy for this process?

You are given a closed vessel at 250 K containing 1 mol of solid species A. The vessel is heated to 40OK K and since no solid or liquid material is visible, you have assumed that all of the material is now present in the gaseous state. The container is flexible and it can be assumed that the pressure is maintained at 1.0 atm throughout this process. Assume ideal gas and ideal solution behavior. You were given the following physical property data for species A Solid-phase heat capacity- 120 J/mol K Liquid-phase heat capacity = 140 J/mol K Gas-phase heat capacity-82 J/mol K Melting point 278 K Latent heat of melting 10 kJ/mol Normal boiling point-353 K Latent heat of vaporization = 30 kJ/mol

Explanation / Answer

Enthalpy change to heat the solid from 250 K to 278 K

H1 = moles x Cps x ( T2-T1)

= 1 mol x 120 J/mol·K x (278 - 250) K

= 3360 J x 1kJ/1000 J

= 3.360 kJ

At 278 K, solid melts and convert into liquid

Enthalpy change during phase change

H2 = moles x latent heat of Melting

= 1 mol x 10 kJ/mol = 10 kJ

Now liquid is heated from 278 K to 353 K

Enthalpy change

H3 = moles x Cpl x (T2-T1)

= 1 mol x 140 J/mol·K x (353 - 278)K

= 10500 J x 1kJ/1000 J

= 10.5 kJ

At 353 K, liquid convert into gaseous phase.

Enthalpy change for phase change

H4 = moles x latenr heat of vaporization

= 1 mol x 30 kJ/mol = 30 kJ

This gas is heated from 353 K to 400 K

Enthalpy change

H5 = moles x Cpg x (T2-T1)

= 1 mol x 82 J/mol·K x (400 - 353)K

= 3854 J x 1kJ/1000 J

= 3.854 kJ

Change in enthalpy for the process

= H1 + H2 + H3 + H4 + H5

= 3.360 + 10 + 10.5 + 30 + 3.854

= 57.714 kJ

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