Question

$3$ moles of an ideal gas at a temperature of $27^{\circ} C$ are mixed with $2$ moles of an ideal gas at a temperature $227^{\circ} C$, determine the equilibrium temperature of the mixture, assuming no loss of energy

• A. $327^{\circ} C$
• B. $107^{\circ} C$
• C. $318^{\circ} C$
• D. $410^{\circ} C$

Answer 1

Answer: (B)

Energy possessed by the ideal gas at $27^{\circ} C$ is
$E_{1}=3\left(\frac{3}{2} R \times 300\right)$
$=\left(\frac{2700 R}{2}\right)$
Energy possessed by the ideal gas at $227^{\circ} C$ is
$E_{2}=2\left(\frac{3 R}{2} \times 500\right)=1500\, R$
If $T$ be the equilibrium temperature, of the mixture, then its energy will be
$E_{2}=5\left(\frac{3 R T}{2}\right)$
Since, energy remains conserved,
$E_{m}=E_{1}+E_{2}$
or $5\left(\frac{3 R T}{2}\right)=\frac{2700 R}{2}+1500\, R$
or $T=380\, K$ or $107^{\circ} C$