Question

A 3.65-mol sample of an ideal diatomic gas expands adiabatically from a

volume of 0.1210 m 3 to 0.750 m 3 . Initially the pressure was 1.00 atm. Determine:

a) the initial and final temperatures, [Ans = 404 K, 195K]

b) the change in internal energy, [Ans = −1.59 X 10 4 J]

c) the heat lost by the gas, [Ans = take a guess]

d) the work done on the gas [Ans = having guessed part c) right, you should…]

Note. 1.00 atm = 101.3 kPa. Assume no molecular vibration but still use

diatomic adjustment values.

Answer 1

Given:

$n=3.65\:\rm mol$

$V_1=0.121\:\rm m^3$

$V_2=0.750\:\rm m^3$

$P_1=\rm 1\: atm=101.3*10^3\: Pa$

$\gamma=1.4$

(a)

$$PV=nRT$$

$$T_1=\frac{P_1V_1}{nR}=\frac{101.3*10^3*0.121}{3.65*8.31}=404\:\rm K$$

$$TV^{\gamma-1}=const$$

$$T_2=T_1(V_1/V_2)^{\gamma-1}=404(0.121/0.750)^{0.4}=195\:\rm K$$

(b)

$$\Delta U=-W=-\frac{P_1V_1^{\gamma}}{1-\gamma}(V_2^{1-\gamma}-V_1^{1-\gamma})$$

$$\Delta U=\frac{101.3*10^3*0.121^{1.4}}{0.4}(0.750^{-0.4}-0.121^{-0.4})\\=-1.59*10^4\:\rm J$$

(c) adiabatic process is a process without heat exchange, hence

$$Q=0$$

(d) 

$$W=-\Delta U=1.59*10^4\:\rm J$$