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Standard XII

Chemistry

Ideal Gas Equation

A rigid and i...

Question

A rigid and insulated tank of 3 $m^{3}$ volume is divided into two components. One compartment of volume of 2 $m^{3}$ contains an ideal gas at 0.8314 MPa and 400 K and while the second compartment of volume 1 $m^{3}$ contains the same gas 8.314 MPa and 500 k. If the partition between the two compartments is ruptured, the final temperature of the gas is:

420 K

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450 K

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480 K

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None of these

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Solution

The correct option is A 420 K
First compartment
$P_{1} = 0.8314 M P a V_{1} = 2 x m^{2} T_{1} = 400 K$
Second compartment
$P_{2} = 8.314 M P a V_{2} = 1 m^{3} T_{2} = 500 K$
$n_{1} = \frac{P_{1} V_{1}}{R T_{1}} = \frac{0.8314 \times 10^{6} \times 2}{8.314 \times 400} = 500$
$n_{2} = \frac{P_{2} V_{2}}{R T_{2}} = \frac{8.314 \times 1 \times 10^{6}}{8.314 \times 500} = 200$
After Ruptewe of compartment
$n = 500 + 200 = 700 V = 2 P_{1} = 3 m^{3}$
$P = 0.8314 M P q$
$T = \frac{P V}{n R} = \frac{3 \times 8.314 \times 10^{6}}{700 \times 8.314}$
$= \frac{3000}{7}$
$≃ 420 K$

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