Discussion Forum

1)

A sample of argon of 1 atm pressure and 300 K expands reversibly and adiabatically from $1.25 dm^{3}$ to $2.5dm^{3}$ . Calculate  the approximate enthalpy (in J) change

(i) $C_{v}$ for argon is $12.48 JK^{-1}$

(ii) Assume argon to be an ideal gas

(iii) $\triangle T=111.5K$

Answer:

Option B

Explanation:

 We know that, $\triangle H=nC_{p} \triangle T$

 Number of moles of argon (n)= $\frac{pV}{RT}$

 given, p=1 atm, V=1.25 dm³, T=300 K

 $\therefore$   $n= \frac{1 \times 1.25}{0.082 \times 300}$=0.05 mol

 also, $C_{p}C_{v}=12.48 JK^{-1}$

 $C_{p}=C_{v}+R=12.48+8.314=20.794$

 On substituting  the above values in Eq.(i) ,

 we get

 $\triangle H=0.05 \times 20.794 \times 111.5=115.92$ ≈ 117 J