First slide

Heat ;work and internal energy

Question

A closed vessel contains 0.1 mole of a monatomic ideal gas at 200 K. If 0.05 mole of the same gas at 400 K is added to it, the final equilibrium temperature (in K) of the gas in the vessel will be close to ___________ .

Moderate

Solution

Assuming $Q = 0, W = 0$
We have $Δ U = 0$
$\Rightarrow n_{1} C_{V_{1}} (T_{f} - T_{1}) + n_{2} C_{V_{2}} (T_{f} - T_{2}) = 0$
$\Rightarrow n_{1} C_{V_{1}} (T_{f} - T_{1}) = - n_{2} C_{V_{2}} (T_{f} - T_{2}) = 0$
$\Rightarrow n_{1} (T_{f} - T_{1}) = - n_{2} (T_{f} - T_{2}) (∵ F o r s a m e g a s C_{V_{1}} = C_{V_{2}})$
$\Rightarrow n_{1} T_{f} - n_{1} T_{1} = - n_{2} T_{f} + n_{2} T_{2}$
$\Rightarrow n_{1} T_{f} + n_{2} T_{f} = n_{1} T_{1} + n_{2} T_{2}$

$\Rightarrow T_{f} = \frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}} = \frac{(0.1) (200) + (0.05) (400)}{0.1 + 0.05}$
$= \frac{20 + 20}{0.15} = 266.6666 K$

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