A sample of argon gas at 1 atm pressure and 27° C expands reversibly and adiabatically from 1.25 dm 3 to 2.50 dm 3. Calculate the enthalpy changc in this process CVm for argon is 12.49 JK -1 mol -1.

Question

A sample of argon gas at 1 atm pressure and 27° C expands reversibly and adiabatically from 1.25 dm 3 to 2.50 dm 3. Calculate the enthalpy changc in this process CVm for argon is 12.49 JK -1 mol -1. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Given: CV=12.49 ⇒ Cp=20.8 ⇒ (Cp/CV)=&gamma;=1.66 In case of reversible adiabatic expansion: T V&gamma;-1= constant ⇒ (T2/T1)=((V1/V2))&gamma;-1=((V1/V2))0.66 ⇒ T2=T1((V1/V2))0.66 ⇒ Δ H=n Cp Δ T =(1 × 1.25/0.082 × 300) × 20.8 ×(189.86-300) J =-116.4 J

Q. A sample of argon gas at 1 atm pressure and 27∘C expands reversibly and adiabatically from 1.25dm3 to 2.50dm3. Calculate the enthalpy changc in this process CVm​​ for argon is 12.49JK−1mol−1.

Given : CV​=12.49⇒Cp​=20.8 ⇒CV​Cp​​=γ=1.66 In case of reversible adiabatic expansion : TVγ−1= constant ⇒T1​T2​​=(V2​V1​​)γ−1=(V2​V1​​)0.66 ⇒T2​=T1​(V2​V1​​)0.66 ⇒ΔH=nCp​ΔT =0.082×3001×1.25​×20.8×(189.86−300)J =−116.4J

Q. A sample of argon gas at $1$ atm pressure and $2 7^{\circ} C$ expands reversibly and adiabatically from $1.25 d m^{3}$ to $2.50 d m^{3}$ . Calculate the enthalpy changc in this process $C_{V_{m}}$ for argon is $12.49 J K^{- 1} m o l^{- 1}$ .