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Standard XII

Chemistry

Heat of Reaction

A rigid and i...

Question

A rigid and insulated tank of $3 m^{3}$ volume is divided into two compartments. One compartment of volume of $2 m^{3}$ 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:

422.8 K

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452.5 K

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480.0 K

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492.3 K

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Solution

The correct option is D 492.3 K
Mole of the gas in the first compartment
$n_{1} = \frac{P_{1} V_{1}}{R T_{1}} = - \frac{5 \times 10^{6} \times 2}{25 / 3 \times 400} = 3000$
$n_{2} = - \frac{50 \times 10^{6} \times 2}{25 / 3 \times 500} = 24000$
The tank is rigid and insulated hence w = 0
and q = 0 therefore $Δ U = 0$
Let $T_{f} a n d P_{f}$ denote the final temperature and pressure respectively $Δ U = n_{1} C_{v, m} [T_{f} - T_{1}] + n_{2} C_{v, m} [T_{f} - T_{2}]$
$= 0$
$2000 (T_{f} - 400) + 24000 (T_{f} - 500) = 0$
$T_{f} = 492.3 K$

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