216

You visited us 216 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Ideal Gas Equation

In a 2.5 L ...

Question

In a 2.5 $L$ flask at $27^{\circ} C$ temperature, the pressure of a gas was found to be 8 $a t m$ . If $2.41 \times 10^{23}$ molecules of the same gas are introduced into the container, the temperature changed to $T_{2}$ . The pressure of gas is found to be 10 $a t m$ . Find out the value of $T_{2}$ :

253

K

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

347

K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

230

K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

370

K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 253 $K$
$6.023 \times 10^{23}$ molecules corresponds to 1 $m o l e$
$2.41 \times 10^{23}$ molecules corresponds to $= \frac{2.41 \times 10^{23}}{6.023 \times 10^{23}} = 0.4 m o l e s$
Case (i)
$P_{1} V_{1} = n_{1} R_{1} T_{1}$
$8 (2.5) = n_{1} (0.08) (300)$
$n_{1} = \frac{8 (2.5)}{(0.08) (300)} = 0.833 m o l e s$
Total moles $n_{2} = 0.833 + 0.4$
$n_{2} = 1.233$
Case (ii)
$P_{2} V_{2} = n_{2} R T_{2}$
$10 (2.5) = 1.233 (0.08) T_{2}$
$T_{2} = 253.4$

Suggest Corrections

Similar questions

Q. When

2 g

of a gas A is introduced into an evacuated flask kept at

25^{\circ} C

, the pressure is found to be

1 a t m

. If

3 g

of another gas B is then added to the same flask at the same temperature, the total pressure becomes

1.5 a t m

. Assuming ideal gas behaviour, the value of

\frac{M_{B}}{M_{A}}

to the nearest integer is :

Q. When

2 g

of a gas A is introduced into an evacuated flask kept at

25^{\circ} C

, the pressure is found to be

1 a t m

. If

3 g

of another gas B is then added to the same flask at the same temperature, the total pressure becomes

1.5 a t m

. Assuming ideal gas behaviour, the value of

\frac{M_{B}}{M_{A}}

to the nearest integer is :

Q. Fixed mass of an ideal gas at

290

K and

1

atm pressure occupied a volume of

0.2 m^{2}

. Calculate the volume of the gas if pressure and temperature and changed to

1.2159 \times 10^{5}

Pa and

253

k respectively.

Q. 4 g of an ideal gas was introduced into a bulb of volume of

0.821 {dm}^{3}

at certain pressure, P and temperature T. The above bulb was placed in a thermostat maintained at temperature (T+125)K. 0.8 gm of the gas was leaked to keep the original pressure. Calculate the pressure in atm. [Molar mass of the gas is

40 {gmol}^{- 1}

and R value is

0.0821 {lit-atm-k}^{- 1} {mol}^{- 1}

]

Q. 28 G of N2 gas is contained in a flask at a pressure of 10 atm and at a temperature of

57^{\circ} C

. It is found that due to leakage in the flask, the pressure is reduced to half and the temperature to

27^{\circ} C

. The quantity of

N_{2}

gas that leaked out is

Join BYJU'S Learning Program

In a 2.5 L flask at 27∘C temperature, the pressure of a gas was found to be 8 atm. If 2.41×1023 molecules of the same gas are introduced into the container, the temperature changed to T2. The pressure of gas is found to be 10 atm. Find out the value of T2:

The correct option is B 253 K6.023×1023 molecules corresponds to 1 mole2.41×1023 molecules corresponds to =2.41×10236.023×1023=0.4 molesCase (i)P1V1=n1R1T18(2.5)=n1(0.08)(300)n1=8(2.5)(0.08)(300)=0.833 molesTotal moles n2=0.833+0.4n2=1.233Case (ii)P2V2=n2RT210(2.5)=1.233(0.08)T2T2=253.4