Question

) A 3.65-mol sample of an ideal diatomic gas expands adiabatically from a volume of 0.1210 m3 to 0.750 m3. Initially the pressure was 1.00 atm. Determine: (a) the initial and final temperatures; (b) the change in internal energy; (c) the heat lost by the gas; (d) the work done on the gas. (Assume no molecular vibration.)

Answer 1

Solution:-

(a)

$T_1 = {P_1 V_1\over nR}$

$T_1 = {0.1210*1\over 3.65*0.082}$

T₁= 0.390008058

And

P₁V₁^7/5 = P₂V₂^7/5

^P₂ = 0.777

$T_2= {P_2V_2\over nR}$ = $0.77*0.750\over 3.65*0.082$

=1.92950217

(b)$\Delta U =({ 5\over 2})*1*(T_2-T_1)$ = 3.75

(c)

Heat loss $\implies$ $\Delta Q =0$

(d)

$W = - \Delta U$

= -3.75