A balloon containing 1 mole air at 1 atm initially is filled further with air till pressure increases to 4 atm. The initial diameter of the balloon is 1 m and the pressure at each stage is proportion to diameter of the ballon. How many no. of moles of air added to change the pressure from 1 atm to 4 atm

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257

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255

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256

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Solution

The correct option is C
255

$P \propto d$ P = kd and $k = \frac{1 a t m}{1 m e t r e}$

$P V = n R T$ $k d (\frac{1}{6} π d^{3})$ = $n R T$ ;

$\frac{d_{1}^{4}}{d_{2}^{4}}$ = $\frac{n_{1}}{n_{2}}$ ; $\frac{1}{4^{4}}$ = $n_{1} / n_{2}$ ; $n_{2} = 256$

no. of moles added = 256 - 1 = 255

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Similar questions

Q. A balloon containing 1 mole of air at 1 atm initially, is filled further till pressure increases to 4 atm. The initial diameter of the balloon is 1 m and the pressure at each stage is proportional to diameter of the balloon. How many moles of air are added to change the pressure from 1 atm to 4 atm?

Q. A balloon containing 1-mole air at

1 a t m

initially is filled further with air till pressure increases to

4 a t m

. The initial diameter of the balloon is

1 m

and the pressure at each stage is proportional to the diameter of the balloon. How many no. of moles of air added to change the pressure from

1 a t m

4 a t m

Q. A balloon containing 1 mol of air initiallyat 1 atm is filled further with air till the pressure increases to 4 atm. The initial diameter of the balloon is

1 m

and the pressure at each stage is proportional to diameter of the balloon. How many moles of air was to change the pressure from 1 atm to 4 atm?

A balloon containing 1 mole air at 1 atm initially is filled further with air till pressure increases to 4 atm. The initial diameter of the balloon is 1 m and the pressure at each stage is proportion to diameter of the ballon. How many moles of air have to be added to change the pressure from 1 atm to 4 atm

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mole air at

1

atm initially is filled further with air till pressure increases to

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The correct option is C 255 P∝d P = kd and k=1 atm1 metre PV=nRT kd(16πd3) = nRT; d41d42=n1n2; 144=n1/n2; n2=256 no. of moles added = 256 - 1 = 255

The correct option is C
255

$P \propto d$ P = kd and $k = \frac{1 a t m}{1 m e t r e}$

$P V = n R T$ $k d (\frac{1}{6} π d^{3})$ = $n R T$ ;

$\frac{d_{1}^{4}}{d_{2}^{4}}$ = $\frac{n_{1}}{n_{2}}$ ; $\frac{1}{4^{4}}$ = $n_{1} / n_{2}$ ; $n_{2} = 256$

no. of moles added = 256 - 1 = 255