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3 moles of an ideal gas at a temperature of $27^{\circ} \mathrm{C}$ are mixed with 2 moles of an ideal gas at a temperature $227^{\circ} \mathrm{C}$, determine the equilibrium temperature of the mixture, assuming no loss of energy.
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Verified Answer
The correct answer is:
$107^{\circ} \mathrm{C}$
Energy possessed by the ideal gas at $27^{\circ} \mathrm{C}$ is
$$
E_{1}=3\left(\frac{3}{2} R \times 300\right)=\frac{2700 R}{2}
$$
Energy possessed by the ideal gas at $227^{\circ} \mathrm{C}$ is
$$
\mathrm{E}_{2}=2\left(\frac{3 \mathrm{R}}{2} \times 500\right)=1500 \mathrm{R}
$$
If $\mathrm{T}$ be the equilibrium temperature, of the mixture, then its energy will be
$$
\mathrm{E}_{\mathrm{m}}=5\left(\frac{3 \mathrm{R} \mathrm{T}}{2}\right)
$$
Since, energy remains conserved,
$$
\mathrm{E}_{\mathrm{m}}=\mathrm{E}_{1}+\mathrm{E}_{2}
$$
or $\quad 5\left(\frac{3 R T}{2}\right)=\frac{2700 R}{2}+1500 R$
or $\quad \mathrm{T}=380 \mathrm{~K}$ or $107^{\circ} \mathrm{C}$
$$
E_{1}=3\left(\frac{3}{2} R \times 300\right)=\frac{2700 R}{2}
$$
Energy possessed by the ideal gas at $227^{\circ} \mathrm{C}$ is
$$
\mathrm{E}_{2}=2\left(\frac{3 \mathrm{R}}{2} \times 500\right)=1500 \mathrm{R}
$$
If $\mathrm{T}$ be the equilibrium temperature, of the mixture, then its energy will be
$$
\mathrm{E}_{\mathrm{m}}=5\left(\frac{3 \mathrm{R} \mathrm{T}}{2}\right)
$$
Since, energy remains conserved,
$$
\mathrm{E}_{\mathrm{m}}=\mathrm{E}_{1}+\mathrm{E}_{2}
$$
or $\quad 5\left(\frac{3 R T}{2}\right)=\frac{2700 R}{2}+1500 R$
or $\quad \mathrm{T}=380 \mathrm{~K}$ or $107^{\circ} \mathrm{C}$
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