JEE MAIN - Chemistry (2019 - 11th January Morning Slot - No. 16)
Two blocks of the same metal having same mass and at temperature T1 and T2, respectively, are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy, $$\Delta $$S, for this process is :
2Cp In $$\left[ {{{{{\left( {{T_1} + {T_2}} \right)}^{{1 \over 2}}}} \over {{T_1}{T_2}}}} \right]$$
2Cp In $$\left[ {{{\left( {{T_1} + {T_2}} \right)} \over {2{T_1}{T_2}}}} \right]$$
Cp In $$\left[ {{{{{\left( {{T_1} + {T_2}} \right)}^2}} \over {4{T_1}{T_2}}}} \right]$$
2Cp In $$\left[ {{{\left( {{T_1} + {T_2}} \right)} \over {4{T_1}{T_2}}}} \right]$$
Explanation
When two blocks comes in contact with each other and attain thermal equilibrium then
final temperature of the blocks,
Tf = $${{{T_1} + {T_2}} \over 2}$$
$$\Delta $$S = $$\Delta $$S1 + $$\Delta $$S2
= Cp $$\ln \left( {{{{T_f}} \over {{T_1}}}} \right)$$ + Cp $$\ln \left( {{{{T_f}} \over {{T_2}}}} \right)$$
= Cp $$\ln \left( {{{{T_1} + {T_2}} \over {2{T_1}}}} \right)$$ + Cp $$\ln \left( {{{{T_1} + {T_2}} \over {2{T_2}}}} \right)$$
= Cp $$\ln \left( {{{{{\left( {{T_1} + {T_2}} \right)}^2}} \over {4{T_1}{T_2}}}} \right)$$
final temperature of the blocks,
Tf = $${{{T_1} + {T_2}} \over 2}$$
$$\Delta $$S = $$\Delta $$S1 + $$\Delta $$S2
= Cp $$\ln \left( {{{{T_f}} \over {{T_1}}}} \right)$$ + Cp $$\ln \left( {{{{T_f}} \over {{T_2}}}} \right)$$
= Cp $$\ln \left( {{{{T_1} + {T_2}} \over {2{T_1}}}} \right)$$ + Cp $$\ln \left( {{{{T_1} + {T_2}} \over {2{T_2}}}} \right)$$
= Cp $$\ln \left( {{{{{\left( {{T_1} + {T_2}} \right)}^2}} \over {4{T_1}{T_2}}}} \right)$$
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