Question

A balloon filled with methane is pricked with a sharp point and quickly plunged into a tank of hydrogen at the same pressure.After sometime,the balloon will have
(a)Enlarged
(b)Shrinked
(c)Remain unchanged in size
(d)Ethylene inside it

Answer 1

Answer: The balloon will be enlarged.

Explanation: It can be explained by Graham's Law of effusion.

Graham's Law: This law states that the rates of effusion of gases is inversely proportional to the square roots of their molar masses at the same temperature and pressure.

$\frac{Rate_A}{Rate_B}=\sqrt{\frac{MolarMass_B}{MolarMass_A}}$                  ......(1)

where,

$Rate_A$ = rate of gas A.

$Rate_B$ = Rate of gas B

In the question, it is given that the balloon was filled with methane first and then it was plunged into the tank filled with hydrogen gas, we take the rate of Hydrogen to rate of methane, we get

Gas A = Hydrogen gas

Gas B = Methane gas

Molar mass of Hydrogen gas = 2g/mol

Molar mass of Methane gas = 16g/mol

Putting in the values in equation 1, we get

$\frac{Rate(H_2)}{Rate(CH_4)}=\sqrt{\frac{16}{2}}$

$Rate(H_2)=\frac{4}{\sqrt{2}}\times Rate(CH_4)$

$Rate(H_2)=2.828\times Rate(CH_4)$

Hence, the balloon will be enlarged.

Answer 2

Answer:

$\textbf{Enlarge}$

• According to the relation :

Rate of Diffusion of Gas A ÷ Rate of Diffusion of Gas B = √ Molar mass of Gas B ÷ Molar mass of Gas A.

• $\textbf{ Solution :}$

=> R(H2) / R(CH4) = √16/2

=> R(H2 = 2√2 × R(CH4)

Hope it helps!!

@charlie16.