Question 542: Thermodynamics & KTG

Question

Two thermally insulated vessels $$1$$ and $$2$$ are filled with air at temperatures $$\left( {{T_1},{T_2}} \right),$$ volume $$\left( {{V_1},{V_2}} \right)$$ and pressure $$\left( {{P_1},{P_2}} \right)$$ respectively. If the value joining the two vessels is opened, the temperature inside the vessel at equilibrium will be

Accepted Answer

Here $$Q=0$$ and $$W=0.$$ Therefore from first law of thermodynamics $$\Delta U = Q + W = 0$$
$$	herefore$$ Internal energy of the system with partition $$=$$ Internal energy of the system without partition.
$${n_1}{C_v}\,{T_1} + {n_2}\,{C_v}{T_2} = \left( {{n_1} + {n_2}} ight){C_v}\,T$$
$$	herefore$$ $$T = {{{n_1}{T_1} + {n_2}T{}_2} \over {{n_1} + {n_2}}}$$
But $${n_1} = {{{P_1}{V_1}} \over {R{T_1}}}$$ and $${n_2} = {{{P_2}{V_2}} \over {R{T_2}}}$$
$$	herefore$$ $$T = {{{{{P_1}{V_1}} \over {R{T_1}}} 	imes {T_1} + {{{P_2}{V_2}} \over {R{T_2}}} 	imes {T_2}} \over {{{{P_1}{V_1}} \over {R{T_1}}} + {{{P_2}{V_2}} \over {R{T_2}}}}}$$
$$ = {{{T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} ight)} \over {{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}}}$$

Answer

$${T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)/\left( {{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}} \right)$$

Answer

$$\left( {{T_1} + {T_2}} \right)/2$$

Answer

$${{T_1} + {T_2}}$$

Answer

$${T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)/\left( {{P_1}{V_1}{T_1} + {P_2}{V_2}{T_2}} \right)$$

Thermodynamics & KTG

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