Two moles of an ideal monatomic gas are initially at 300 K and 1 bar. The gas is
ID: 479276 • Letter: T
Question
Two moles of an ideal monatomic gas are initially at 300 K and 1 bar. The gas is compressed to a final pressure of 2.5 bar in a piston. Calculate V_i, v_f, T_f, q, w, delta U, and delta H for each of the following situations. The gas is compressed reversibly and isothermally The gas is compressed reversibly and adiabatically. The gas is compressed irreversibly and adiabatically when 2.5 bar is applied instantaneously in a single step. The piston oscillates wildly but finally comes to rest at P_f = 2.5 bar.Explanation / Answer
(a)
Initial volume = nRT/P = 2 x 8.314 x 300 /105 Pa == 0.05 m3
Isothermal process. So, final temperature = 300 K
Final volume is given by equation,
P1V1 =P2V2
1 bar x 0.05 m3 = 2.5 bar x Vf
Vf = 0.02 m3
Iosthermal process, delta H and delta U = 0
work done = -Pi (Vf-Vi) == - 105 Pa x(0.02 -0.05)
= 3000 J
heat = -w = -3000 J
(b)
in adiabatic process, relation between P and T is,
P1-y Ty = constant
Gamma for monoatomic gas = 1.66
(1)1-1.66 x 3001.66 = (2.5)1-1.66 x Tf1.66
Tf = 208.4 K
In adiabatic process, q =0
for finding final volume use this equation,
Initial volume = nRT/P = 2 x 8.314 x 300 /105 Pa == 0.05 m3
Final volume = P1V1y / P2 == (1 x 0.051.66 ) / 2.5
= 0.00277 m3
work done = R (T2-T1) /(1-y)
= 8.314 (208.4 K - 300 K ) / (1-1.66)
= 1154 J
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