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  4. At constant temperature, the volume of a gas is to be decreased by 4%. The pressure must be increased by

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Q. At constant temperature, the volume of a gas is to be decreased by 4%. The pressure must be increased by

Physical World, Units and Measurements

Solution:

At constant temperature, $ p_1 V_1 = p_2 V_2 $
$\Rightarrow \frac{ p_1 }{ p_2 } = \frac{ V_2 }{ V_1 } $
Hence fractional change in volume
$\Rightarrow \frac{ V_1 - V_2 }{ V_1 } = \frac{ 4 }{ 100} = \frac{ 1}{ 25 } $
$\Rightarrow 1 - \frac{ V_2 }{ V_1 } = \frac{ 1}{25} $
$\Rightarrow \frac{ V_2 }{ V_1 } = \frac{ 24}{25} $
$\Rightarrow \frac{ p_1 }{ p_2 } = \frac{ V_2 }{ V_1 } = \frac{ 24}{25} $
$\Rightarrow \frac{ p_2 - p_! }{ p_1 } = \frac{ 24}{25} - 1 = \frac{ 1}{ 24 }$
Percentage increase in pressure = $ \frac{ 100}{ 24} = $ 4.16%.