Thermodynamics Question 6
Question 6 - 30 January - Shift 2
1 mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27^{\circ} C$. The work done is $3 kJ mol^{-1}$. The final temperature of the gas is $K$ (Nearest integer). Given $C_v=20 J mol^{-1} K^{-1}$.
Show Answer
Answer: (150)
Solution:
Formula: Gibb’s Free Energy
$ q=0 $
$\Delta U=w$
$1 \times 20 \times[T_2-300]=-3000$
$T_2-300=-150$
$T_2=150 K$
