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Standard XII

Chemistry

Real Gases

Using Vander ...

Question

Using Vander Waals equation, calculate the constant "a" when 2 moles of a gas confined in a 4 litre flask exert a pressure of 11.0 atm at a temperature of 300K. The value of "b" is 0.05 litre ${m o l}^{- 1}$

4.25

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6.96

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7.26

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Solution

The correct option is B 6.96
$(p + \frac{n^{2} a}{v^{2}}) (v - n b) = n R T$

$v = 42 p = 11 a t m T = 300 k$

$b = 0.05 l / m o l n = 2$

$(11 + \frac{2^{2} a}{4^{2}}) (4 - 2 \times 0.05) = 2 \times 0.082 \times 300$

$a = 6.96 a t m l^{2} m o l^{- 2}$

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Using Vander Waals equation, calculate the constant "a" when 2 moles of a gas confined in a 4 litre flask exert a pressure of 11.0 atm at a temperature of 300K. The value of "b" is 0.05 litre mol−1

The correct option is B 6.96(p+n2av2)(v−nb)=nRTv=42p=11atmT=300kb=0.05l/moln=2(11+22a42)(4−2×0.05)=2×0.082×300a=6.96atml2mol−2

Using Vander Waals equation, calculate the constant "a" when 2 moles of a gas confined in a 4 litre flask exert a pressure of 11.0 atm at a temperature of 300K. The value of "b" is 0.05 litre ${m o l}^{- 1}$

The correct option is B 6.96
$(p + \frac{n^{2} a}{v^{2}}) (v - n b) = n R T$

$v = 42 p = 11 a t m T = 300 k$

$b = 0.05 l / m o l n = 2$

$(11 + \frac{2^{2} a}{4^{2}}) (4 - 2 \times 0.05) = 2 \times 0.082 \times 300$

$a = 6.96 a t m l^{2} m o l^{- 2}$