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Question Number 50275 by Tawa1 last updated on 15/Dec/18

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Dec/18

△S=∫_T_1 ^T_2 C_v (dT/T)−R∫_V_1 ^V_2 (dV/V) V_1 =1×10^(−3) m^3 V_2 =3×10^(−3) m^3 R=8.314 T_1 =100K T_2 =400K C_v =45+6×10^(−3) T+8×10^(−6) T^2 △S=∫_T_1 ^T_2 (a+bT+cT^2 )(dT/T)−R∫_V_1 ^V_2 (dV/V) =∣alnT+bT+c×(T^2 /2)∣_T_1 ^T_2 −R∣lnV∣_V_1 ^V_2 =aln((T_2 /T_1 ))+b(T_2 −T_1 )+(c/2)(T_2 ^2 −T_1 ^2 )−Rln((V_2 /V_1 )) =45ln(4)+6×10^(−3) ×300+4×10^(−6) ×15×10^4 −8.314ln3 =45ln4+1.8+0.6−8.324ln3 =2.4+45ln4−8.324ln3 ≈55.64

$$\bigtriangleup{S}=\int_{{T}_{\mathrm{1}} } ^{{T}_{\mathrm{2}} } {C}_{{v}} \frac{{dT}}{{T}}−{R}\int_{{V}_{\mathrm{1}} } ^{{V}_{\mathrm{2}} } \frac{{dV}}{{V}} \\ $$$${V}_{\mathrm{1}} =\mathrm{1}×\mathrm{10}^{−\mathrm{3}} {m}^{\mathrm{3}} \:\:\:\:{V}_{\mathrm{2}} =\mathrm{3}×\mathrm{10}^{−\mathrm{3}} {m}^{\mathrm{3}} \\ $$$${R}=\mathrm{8}.\mathrm{314} \\ $$$${T}_{\mathrm{1}} =\mathrm{100}{K}\:\:\:{T}_{\mathrm{2}} =\mathrm{400}{K} \\ $$$${C}_{{v}} =\mathrm{45}+\mathrm{6}×\mathrm{10}^{−\mathrm{3}} {T}+\mathrm{8}×\mathrm{10}^{−\mathrm{6}} {T}^{\mathrm{2}} \\ $$$$\bigtriangleup{S}=\int_{{T}_{\mathrm{1}} } ^{{T}_{\mathrm{2}} } \left({a}+{bT}+{cT}^{\mathrm{2}} \right)\frac{{dT}}{{T}}−{R}\int_{{V}_{\mathrm{1}} } ^{{V}_{\mathrm{2}} } \frac{{dV}}{{V}} \\ $$$$=\mid{alnT}+{bT}+{c}×\frac{{T}^{\mathrm{2}} }{\mathrm{2}}\mid_{{T}_{\mathrm{1}} } ^{{T}_{\mathrm{2}} } −{R}\mid{lnV}\mid_{{V}_{\mathrm{1}} } ^{{V}_{\mathrm{2}} } \\ $$$$={aln}\left(\frac{{T}_{\mathrm{2}} }{{T}_{\mathrm{1}} }\right)+{b}\left({T}_{\mathrm{2}} −{T}_{\mathrm{1}} \right)+\frac{{c}}{\mathrm{2}}\left({T}_{\mathrm{2}} ^{\mathrm{2}} −{T}_{\mathrm{1}} ^{\mathrm{2}} \right)−{Rln}\left(\frac{{V}_{\mathrm{2}} }{{V}_{\mathrm{1}} }\right) \\ $$$$=\mathrm{45}{ln}\left(\mathrm{4}\right)+\mathrm{6}×\mathrm{10}^{−\mathrm{3}} ×\mathrm{300}+\mathrm{4}×\mathrm{10}^{−\mathrm{6}} ×\mathrm{15}×\mathrm{10}^{\mathrm{4}} −\mathrm{8}.\mathrm{314}{ln}\mathrm{3} \\ $$$$=\mathrm{45}{ln}\mathrm{4}+\mathrm{1}.\mathrm{8}+\mathrm{0}.\mathrm{6}−\mathrm{8}.\mathrm{324}{ln}\mathrm{3} \\ $$$$=\mathrm{2}.\mathrm{4}+\mathrm{45}{ln}\mathrm{4}−\mathrm{8}.\mathrm{324}{ln}\mathrm{3} \\ $$$$\approx\mathrm{55}.\mathrm{64} \\ $$

Commented by Tawa1 last updated on 15/Dec/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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