The enthalpy of formation for CO2 (s) and CO2 (g) is: -427.4 KJ/mole and -393.5 KJ/mole, respectively. The sublimation of dry ice is described by CO2 (s) → CO2 (g). The enthalpy needed to sublime 986 grams of CO2 is:
(a) 181.5 Kcal
(b) 611.7 Kcal
(c) 248.3 Kcal
(d) 146.2 Kcal

a. 181.5 kcal

Explanation:

Step 1: Calculate the enthalpy of the process (ΔH°).

Let's consider the following process.

CO₂(s) → CO₂(g)

We can calculate the enthalpy of the process using the following expression.

ΔH° = ∑ np × ΔH°f(p) - ∑ nr × ΔH°f(r)

ΔH° = 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CO₂(s))

ΔH° = 1 mol × (-393.5 kJ/mol) - 1 mol × (-427.4 kJ/mol) = 33.9 kJ

According to the balanced equation, 33.9 kJ are required to sublime 1 mole of CO₂.

Step 2: Convert 986 g of CO₂ to moles

The molar mass of CO₂ is 44.01 g/mol.

986 g × 1 mol/44.01 g = 22.4 mol

Step 3: Calculate the enthalpy needed to sublime 22.4 moles of CO₂

22.4 mol × 33.9 kJ/1 mol = 759 kJ

We can convert it to Kcal using the conversion factor 1 kcal = 4.184 kJ.

759 kJ × 1 kcal/4.184 kJ = 181.5 kcal

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole