A gas occupies a volume of 2.5 L at $9 \times 10^{5} N m^{- 2}$ . Calculate the additional pressure required to decrease the volume of the gas to 1.51. Assume temperature is constant.

15*10⁵ N m^-2

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9*10⁵ N m^-2

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6*10⁵ N m^-2

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24*10⁵ N m^-2

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Solution

The correct option is C
6*10⁵ N m^-2

Here, $P_{1} = 9 \times 10^{5} N m^{- 2}, P_{2} = ?$

$V_{1} = 2.5 L, V_{2} = 1.5 L$

From Boyle's law, $P_{1} V_{1} = P_{2} V_{2}$ , we get

$P_{2} = \frac{P_{1} V_{1}}{V_{2}} = \frac{9 \times 10^{5} \times 2.5}{1.5} = 15 \times 10^{5} N m^{- 2}$

The additional pressure required is $⟮ 15 \times 10^{5} N m^{- 2} ⟯ - ⟮ 9 \times 10^{5} N m^{- 2} ⟯ = 6 \times 10^{5} N m^{- 2}$

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A gas occupies a volume of 2.5 L at 9×105Nm−2. Calculate the additional pressure required to decrease the volume of the gas to 1.51. Assume temperature is constant.

The correct option is C 6*105 N m-2 Here, P1=9×105Nm−2,P2=? V1=2.5 L,V2=1.5 L From Boyle's law, P1V1=P2V2, we get P2=P1V1V2=9×105×2.51.5=15×105Nm−2 The additional pressure required is ⟮15×105Nm−2⟯−⟮9×105Nm−2⟯=6×105Nm−2

A gas occupies a volume of 2.5 L at $9 \times 10^{5} N m^{- 2}$ . Calculate the additional pressure required to decrease the volume of the gas to 1.51. Assume temperature is constant.