Helium gas is filled in two identical bottles A and B. The mass of the gas in the two bottles is 10 gm and 40 gm respectively. If the speed of sound is the same in both bottles, wh... Helium gas is filled in two identical bottles A and B. The mass of the gas in the two bottles is 10 gm and 40 gm respectively. If the speed of sound is the same in both bottles, what conclusions will you draw? Read more

Steps to Solve

1. Understanding the speed of sound in gases The speed of sound in an ideal gas is given by the formula: $$ c = \sqrt{\frac{\gamma RT}{M}} $$ where:

• $c$ is the speed of sound,
• $\gamma$ is the adiabatic index (specific heat ratio),
• $R$ is the universal gas constant,
• $T$ is the absolute temperature (in Kelvin),
• $M$ is the molar mass of the gas.

2. Identifying variables for Bottles A and B In this case, the molar mass $M$ will differ because the amount of gas in each bottle is different: 10 gm for A and 40 gm for B. For helium, the molar mass is approximately 4 gm/mol, which means:

• For Bottle A (10 gm): $$ n_A = \frac{10 \text{ gm}}{4 \text{ gm/mol}} = 2.5 \text{ mol} $$
• For Bottle B (40 gm): $$ n_B = \frac{40 \text{ gm}}{4 \text{ gm/mol}} = 10 \text{ mol} $$

3. Relating mass to temperature Since the speed of sound is the same in both bottles, we can set the equations for each bottle equal to one another: $$ \sqrt{\frac{\gamma R T_A}{M_A}} = \sqrt{\frac{\gamma R T_B}{M_B}} $$ Squaring both sides eliminates the square root: $$ \frac{\gamma R T_A}{M_A} = \frac{\gamma R T_B}{M_B} $$ We can cancel $\gamma R$ (as they are constants) and rearrange for temperatures: $$ \frac{T_A}{M_A} = \frac{T_B}{M_B} $$

4. Substituting known values Using the identified masses:

• For Bottle A: $M_A = 10 \text{ gm}$
• For Bottle B: $M_B = 40 \text{ gm}$

Substituting these values in: $$ \frac{T_A}{10} = \frac{T_B}{40} $$

5. Solving for temperature ratio Cross multiplying to find the relationship: $$ 40 T_A = 10 T_B $$ This simplifies to: $$ T_B = 4 T_A $$