At constant temperature- the pressure of given mass of gas is increased by 15- then the percentage change in its volume will be -nearly-11-1-16-6-18-13-

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Toppr Admin · Accepted Answer

By Boyle-apos-s lawP1V1-P2V2here it is given that P2-1-15P1this gives V2-11-15V1so percentage change in volume - V2-x2212-V1V1 x 100-x2212-13-

At constant temperature, if the pressure of given mass of gas is increased by 15% then the percentage change in its volume will be (nearly)

$11.1$ %
$- 16.6$ %
$18$ %
$- 13$ %

By Boyle's law
$P_{1} V_{1} = P_{2} V_{2}$
here it is given that $P_{2} = 1.15 P_{1}$
this gives $V_{2} = \frac{1}{1.15} V_{1}$
so percentage change in volume = $\frac{V_{2} - V_{1}}{V_{1}}$ x 100
$= - 13$ %

At constant temperature, if the pressure of given mass of gas is increased by 15% then the percentage change in its volume will be (nearly)11.1%−16.6%18%−13%

By Boyle's lawP1V1=P2V2here it is given that P2=1.15P1this gives V2=11.15V1so percentage change in volume = V2−V1V1 x 100=−13%

At constant temperature, if the pressure of given mass of gas is increased by 15% then the percentage change in its volume will be (nearly)

$11.1$ %
$- 16.6$ %
$18$ %
$- 13$ %

By Boyle's law
$P_{1} V_{1} = P_{2} V_{2}$
here it is given that $P_{2} = 1.15 P_{1}$
this gives $V_{2} = \frac{1}{1.15} V_{1}$
so percentage change in volume = $\frac{V_{2} - V_{1}}{V_{1}}$ x 100
$= - 13$ %