If pressure increases by 1 % then percentage decrease of volume of an ideal gas at constant temperature is

Question

If pressure increases by 1 % then percentage decrease of volume of an ideal gas at constant temperature is (A) (100/101) % (B) 1.11 % (C) (101/100) % (D) 10.1 %

Tardigrade Admin · Accepted Answer

P1V1=P2V2 100× V1=101× V2⇒ V2=(100/101)V1 % decrease in volume =(V1 - V2/V1)× 100 =(1 - (100/101))× 100 =(100/101) %

Q. If pressure increases by $1%$ then percentage decrease of volume of an ideal gas at constant temperature is

$\frac{100}{101} %$

$1.11%$

$\frac{101}{100} %$

$10.1%$

$P_{1} V_{1} = P_{2} V_{2}$
$100 \times V_{1} = 101 \times V_{2} \Rightarrow V_{2} = \frac{100}{101} V_{1}$
$%$ decrease in volume $= \frac{V _{1} - V _{2}}{V _{1}} \times 100$
$= (1 - \frac{100}{101}) \times 100$
$= \frac{100}{101} %$

Q. If pressure increases by 1% then percentage decrease of volume of an ideal gas at constant temperature is

101100​%

1.11%

100101​%

10.1%