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Byju's Answer

Standard XII

Chemistry

Graham's Law of Effusion

An open flask...

Question

An open flask has helium gas at $2$ atm and $327^{0}$ C. The flask is heated to $527^{0}$ C at same pressure. The fraction of original gas remaining in the flask is:

\frac{3}{4}

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\frac{1}{4}

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\frac{1}{2}

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\frac{2}{5}

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Solution

The correct option is A $\frac{3}{4}$
Here $P$ & $V$ remain constant

$T_{1} = 327 + 273 = 600 K$
$T_{2} = 527 + 273 = 800 K$

$P V = n R T$
So, $\frac{P V}{R} = n_{1} T_{1} = n_{2} T_{2}$ & $\frac{n_{2}}{n_{1}} = \frac{T_{1}}{T_{2}} = \frac{600}{800}$
$= \frac{3}{4}$ .

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Similar questions

Q. An open flask has Helium gas at 2atm and 327

^{\circ} C

. The flask is heated at 527

^{\circ} C

at the same pressure. The fraction of original gas remaining in the flask is?

Q. A

5 l i t

flask containing

N_{2}

1

bar and

25^{o} C

is connected to a

4 l i t

flask containing

N_{2}

2

bar and

0^{o} C

. After the gases are allowed to mix keeping both flasks at their original temperature, what will be the pressure and amount of

N_{2}

in the

5 l i t

flask assuming ideal gas behaviour.

2 g

of gas

A

introduced in a evacuated flask at

25^{o} C

. The pressure of the gas is

1

atm. Now

3 g

of another gas

B

added to the same flask, the pressure becomes

1.5

atm. The ratio of molecular mass of

A

and

B

2 g

of gas

A

is introduced in an evacuated flask at

25^{\circ}

C. The pressure of the gas is

1

atm. Now

3 g

of another gas

B

is introduced in the same flask and the total pressure becomes

1.5

atm. The ratio of molecular mass of

A

and

B

is:

2 g

of gas

A

introduced in an evacuated flask at

25^{\circ} C

. The pressure of the gas is

1 a t m

. Now,

3 g

of another gas

B

is introduced in the same flask so total pressure becomes

1.5 a t m

. The ratio of the molecular mass of

A

and

B

is:

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An open flask has helium gas at 2 atm and 3270C. The flask is heated to 5270C at same pressure. The fraction of original gas remaining in the flask is:

The correct option is A 34Here P & V remain constantT1=327+273=600KT2=527+273=800KPV=nRTSo, PVR=n1T1=n2T2 & n2n1=T1T2=600800=34 .

An open flask has helium gas at $2$ atm and $327^{0}$ C. The flask is heated to $527^{0}$ C at same pressure. The fraction of original gas remaining in the flask is:

The correct option is A $\frac{3}{4}$
Here $P$ & $V$ remain constant

$T_{1} = 327 + 273 = 600 K$
$T_{2} = 527 + 273 = 800 K$

$P V = n R T$
So, $\frac{P V}{R} = n_{1} T_{1} = n_{2} T_{2}$ & $\frac{n_{2}}{n_{1}} = \frac{T_{1}}{T_{2}} = \frac{600}{800}$
$= \frac{3}{4}$ .