Question

A tank has two rooms separated by a membrane. Room A has 1 kg of air and a volume of 0.5 m3; room B has 0.75 m3 of air with density 0.6 kg/m3. The membrane is broken, and the air comes to a uniform state. Find the final density of the air.

Answer 1

Answer:

The final density of the air is 1.16 kg/m³.

Explanation:

Given that,

Mass of air in room A= 1 kg

Volume of air in room A= 0.5 m³

Volume of air in room B= 0.75 m³

Air density = 0.6 kg/m³

We need to calculate the mass of air in room B

Using formula of density

$\rho=\dfrac{m}{V}$

$m_{B}=\rho\times V$

Put the value into the formula

$m_{B}=0.6\times0.75$

$m_{B}=0.45\ kg$

We need to calculate the combine air mass room A and B

Using formula for total mass

$M=m_{A}+m_{B}$

Put the value into the formula

$M=1+0.45$

$M=1.45\ Kg$

We need to calculate the combine air volume room A and B

Using formula for total mass

$V'=V_{A}+V_{B}$

Put the value into the formula

$V'=0.5+0.75$

$V'=1.25\ Kg$

We need to calculate the final density of the air

Using formula of density

$\rho_{a}=\dfrac{M}{V'}$

Put the value into the formula

$\rho_{a}=\dfrac{1.45}{1.25}$

$\rho_{a}=1.16\ kg/m^3$

Hence, The final density of the air is 1.16 kg/m³.