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Ideal Gas Equation from KTG

A rigid and i...

Question

A rigid and insulated tank of $3 m^{3}$ volume is divided into two compartments. One compartment of volume $2 m^{3}$ contains an ideal gas at $5 MPa$ and $400 K$ and while the second compartment of volume $1 m^{3}$ contains the same gas at $50 MPa$ and $500 K$ . If the partition between the two compartments is ruptured, the final temperature of the gas is:
$(Take: R = \frac{25}{3} {J K}^{- 1} {mol}^{- 1})$

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Solution

Mole of the gas in the first compartment
$n_{1} = \frac{P_{1} V_{1}}{R T_{1}} = \frac{5 \times 10^{6} \times 2}{25 / 3 \times 400} = 3000$
Mole of the gas in the second compartment
$n_{2} = \frac{50 \times 10^{6} \times 1}{25 / 3 \times 500} = 12000$
The tank is rigid and insulated hence $w = 0$
and $q = 0$ therefore $Δ U = 0$
Let $T_{f} and P_{f}$ denote the final temperature and pressure respectively $Δ U = n_{1} C_{v, m} [T_{f} - T_{1}] + n_{2} C_{v, m} [T_{f} - T_{2}]$
$= 0$
$3000 (T_{f} - 400) + 12000 (T_{f} - 500) = 0$
$T_{f} = 480 K$

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