Question

4. What can you tell about the AH of a reaction whose thermogram looks like below? How about Crm of the product related to reactant? Assume that the reference has a Cpm that is independent of temperature. 16 2 Isothermal reversible process A 2.35 mole sample of an ideal gas, for which Cvm 3/2 R initially at 27.0 C and 1.75 x 10 Pa undergoes a transformation. The gas is expanded isothermally and reversibly until the volume triples Fill the table below with values of the final pressure F, as well as q, w, a, and ΔΗ for the described process. Write your answers in the units indicated in the table. Concisely and clearly show your procedure to obtain the answers. 2+2+2+2+2 Δυ Pa

Answer 1

We have a reversible isothermal process. According to the given initial data T₁ = 300 K, P₁ = 17.5 atm (since 1 atm = 10⁵ Pa) , C_v = 3R/2 . Now applying the ideal gas equation for the given sample,

PV = nRT, at const. temp we have P₁ * V₁ = P₂ * V₂ . And we have also been given that the volume gets tripled,

this means that V₂ = 3 * V₁ , Applying the same to the above equation, we get

P₁ = 3 P₂ , This means that   P₂ = P₁ /3 = 17.5/3 = 5.833 atm

Applying the first law of thermodynamics for the given isothermal process, we get

q = U +w

$\Delta$q = $\Delta$U + $\Delta$w

Change in internal energy $\Delta$U = n C_v$\Delta$T = 0, since the temperature is constant

Therefore, $\Delta$q =  $\Delta$w = nRT ln( V₂ / V₁ ) ......[dw= pdV=nRT/V dV , W = nRT/V dV ]

V1

q = w = 2.35 * 8.31 J/mol-K * 300 K * ln (3) = 6436.275 J

Now, we are concerned about $\Delta$H = $\Delta$U + $\Delta$PV

change in enthalpy=change in internal energy + change in PV

Now, since we have considered an ideal gas under consideration and that too isothermal expansion.

$\Delta$PV = 0 (isothermal process), and  $\Delta$U =0 ( already shown above)

Therefore,  $\Delta$H = 0.

So now summarizing all the quantities one by one we get :

P₂ = 5.833 atm

q = 6436.275 J

w = 6436.275 J

$\Delta$U =0

$\Delta$H = 0.