Question

8 g of helium and 4 g of hydrogen gas are present
in a closed vessel at 127°C. If pressure exerted by
the gas mixture is 4.1 atm then the volume of the
container is
[NCERT Pg. 145]
(1) 32 L
(2) 16 L
(3) 40 L
(4) 24L​

Answer 1

$\Large\underline{\textsf{\purple{ \mapsto Required\:Answer:}}}$

Given :- 8 g of helium and 4 g of hydrogen gas are present in a closed vessel at 127°C. If pressure exerted by the gas mixture is 4.1 atm.

To Find :- The volume of thecontainer.

Answer :- Here mass of He = 8g & mass of Hydrogen is 4g . Also we know that Gram Molecular mass of Helium is 4g and that of Hydrogen is 2g . So the number of moles will be ;

By using the formula , $\red{\bf n = \dfrac{Given\:weight}{Molecular\: weight}}$

=> n= 4g/2g .

=> n = 2.

Also ,

=> n' = 8g/4g.

=> n' = 2 .

Now , given that the temperature is 127°C = (127+273)K = 400 Kelvin .( SI unit )

Hence , total number of moles will be = 2+2 = 4 .

Also here , total pressure exerted is 4.1g . And we already know number of moles and temperature .So now on using Ideal Gas Equation , $\red{\bf PV=nRT}$

Where ,

• P is pressure.
• V is volume.
• n is number of moles.
• R is Universal Gas Constant.

$\tt:\implies PV=nRT$

$\tt:\implies 4.1\times V= 4\times 0.082\times 400$

$\tt:\implies V = \dfrac{4\times \cancel{0.082}\times 400}{\cancel{4.1}}$

$\underline{\boxed{\red{\tt:\longmapsto Volume=32L}}}$

★ Hence the volume is 32 L . So option (1) is correct.

Answer 2

Answer:

Option (1)

Explanation:

Consider He 8g/4g = 2 moles

" H 4g/2g = 2 moles

Total no of moles = nHe + nH

2 + 2 = 4 moles

T = 127°C + 273K = 400K

According to ideal gas equation

PV = nRT

we want to find V ?

Therefore, 4.1 x V = 4 x 0.0821 x 400

V = 1600 x 0.0821

4.1

V = 32 L