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valentina_108 [34]
3 years ago
15

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 dens

ity 0.6 kg/m3. The membrane is broken, and the air comes to a uniform state. Find the final density of the air.
Engineering
1 answer:
Mashcka [7]3 years ago
6 0

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.

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