Question

0.08 kg of air at 700 kPa and 800oC is expanded adiabatically to 100 kPa in a closed system.
Taking γ = 1.4 calculate the following.
i. The final temperature.
ii. The work transfer.
iii. The change in internal energy.

Answer 1

a)

$\frac{T_f}{T_i} = \frac{P_i}{P_f}^{\frac{1.4-1}{1.4}} \\ \frac{T_f }{ 800} = \frac{700}{100}^{0.2857} T_f= 1394.87 \;C$

b) no. of moles $= \frac{mass}{ molar \;mass}= \frac{80\; gm }{ 29} = 2.76$

R $= 8.314 \;J /mol \cdot K$

$W= nR \frac{( T_2-T_1) }{ 1- 1.4}= 2.76 \times 8.314 \frac{ 868.02 \;K }{0.4}=4.98 \times 10^4 \;J$

c) change in internal energy = - ve (of work done)