Thermodynamics

Question

0.05 kg of carbon dioxide (molecular weight 44), occupying a volume of 0.03 m³ at 1.025 bar, is compressed reversibly until the pressure is 6.15 bar. Calculate final temperature, the work done on the CO₂, the heat flow to or from the cylinder walls, (i) When the process is according to law pv1.4 = constant, (ii) When the process is isothermal, (iii) When the process takes place in a perfectly thermally insulated cylinder. 

Assume CO₂ to be a perfect gas, and take ϒ = 1.3.

Answer 1

The mass is m = 0.05kg, μ = 44g/mol , the initial volume is,V₁ ​= 0.03m³, the initial pressure is p_1​ = 1.025bar = 1.025⋅10⁵Pa . 

The final pressure is p₂ ​= 6.15 bar = 6.15⋅10⁵Pa.

(i) If pV^1.4 = constant, then \(p_1 V^{1.4}_1 = p_2 V_2^{1.4} \Rightarrow V_2 = V_1(\frac {P_1}{P_2})= 8.3.10^{-3}m^3.\)

For such a polytropic process the work can be calculated as

\(A = \int pdV = v \frac {R}{1.4 -1}.T_1 (1 - \frac {T_2}{T_1}).\)

\(p_1 V_1 =\frac {m}{\mu}RT_1,T_1 = 325.6 K; \,\,p_2 V_2 = \frac {m}{\mu}RT_2, T_2 = 540.5 K.\)

Therefore, the work of gas will be \(A= v \frac {R}{1.4 - 1}.T_1 (1 - \frac {T _2}{T_1})= -5073J.\)

The heat that should be given to gas can be calculated as

\(q = \int T\,ds = v c_v \frac {1.4 - 1.3}{1.4 - 1} (T_2 - T _1) = 3R .v.\frac {1.4 - 1.3}{1.4 - 1}(540.5 - 325 .6) = 1522J.\)

(ii) In the isothermal process T₁ = T₂ = 325.6 K. For such process work can be calculated as \(A= \int p\,dV = vRT \,ln \frac {V_2}{V_1} =vRT \,ln \frac {p_1}{p_2} = -5509 J.\) The change of internal energy is 0, so the heat that should be given to the gas is -5509.J.

(iii) if the cylinder is thermally insulated the heat flow will be 0, so the process will be adiabatic. 

Let us calculate the final temperature:

\(pV ^{1.3} = constant ,\,\, p_1 V^{1.3}_1 =p_2V ^{1.3} _2 \Rightarrow V_2 = V_1 (\frac {p_1}{p_2})^{1/1.3}= 7.56 .10^{-3}m^3.\)

\(p_2 V_2 = \frac {m}{\mu}RT_2, T_2 = 492 .4 K.\)

The work of the gas is equal to

\(A = -(U_2 - U_1)=c_vv(T_1 - T_2)=3Rv (T_1 - T _2) = -4725 J.\)

We may notice that the work of the gas is negative, so we should do work on gas.