Question

15 kg of gas is contained in a cylinder at a pressure of 2 x 107 N/m². The quantity of gas that should be taken out of the cylinder (while keeping temperature constant), such that the pressure becomes 2 x 106 N/m², is A) 10 Kg B) 12 kg C) 13.5 kg D) 15 kg​

Answer 1

Answer:

The amount of gas that is taken out of the cylinder is 13.5kg.

Explanation:

Given:

$m_{1} = 15kg$, $P_{1}=2*10^{7}N/m^{2}$, $P_{2}=2*10^{6}N/m^{2}$

To find:

Mass of gas taken out M

Formula:

P = ρgh

$P=\frac{mgh}{V}$

As the volume and height of the cylinder remain constant, and g is a gravitational constant, we get

P ∝ m

$\frac{P_{1} }{P_{2} }=\frac{m_{1} }{m_{2} }$

$m_{2}=m_{1}\frac{P_{1} }{P_{2} }$

Step 1:

Substituting the given values in the above-derived formula.

$m_{2}=(15)\frac{2*10^{6} }{2*10^{7} }$

$m_{2}= 15 * 10^{-1}$

$\\m_{2}=1.5kg$

$m_{2}$ is the mass of the gas left at pressure $2*10^{6}N/m^{2}$

Step 2:

Mass of gas taken out = Initial mass - Final mass

$M = m_{1}-m_{2}$

$M=15-1.5$

$M=13.5kg$

Therefore, 13.5kg of gas should be taken out of the cylinder so that the pressure becomes $2*10^{6}N/m^{2}$

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Answer 2

Answer:

The correct option is (C).

Explanation:

Given in the question

m1=15kg,P1=2x10^7N/m^2,P2=2x10^6N/m^2

mass of gas taken out M

P=ρgh

P=mgh/V

Volume &height of a cylinder is constant

g=gravitational constant,we get

P∝m

$(P1/P2)=(m1/m2)$

m2=m1x(P2/P1)

putting the values of m1,P1,P2 we get

m2=15x(2x10^6/2x10^7)

m2=15x10^-1

m2=1.5kg

m2=mass of the gas left

mass of gas taken out

      M=m1-m2

      M=15-1.5

      M=13.5kg

So atlast 13.5kg of gas should be taken out of the cylinder so that the pressure becomes 2x10^6.N/m^2