What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 x 103 kg m-3?

As an experienced tutor registered on UrbanPro, I'd like to approach this question by emphasizing the importance of understanding the principles of fluid mechanics and hydrostatics. UrbanPro provides a platform for the best online coaching tuition, where students can delve into such topics with expert guidance.

Now, let's tackle your question. When dealing with the density of water at varying depths under different pressures, we rely on the fundamental concept that pressure increases with depth in a fluid.

The density of water at a depth where the pressure is 80.0 atm can be calculated using the hydrostatic equation:

P=P0+ρghP=P0+ρgh

Where:

• PP is the pressure at the given depth,
• P0P0 is the pressure at the surface (given),
• ρρ is the density of water at the surface (given),
• gg is the acceleration due to gravity (approximately 9.81 m/s29.81m/s2),
• hh is the depth.

Given P0=1.03×103 kg/m3P0=1.03×103kg/m3 (density at the surface) and P=80.0 atmP=80.0atm (pressure at the given depth), we can rearrange the equation to solve for ρρ:

ρ=ρ0⋅(1−P−P0ρ0⋅g)ρ=ρ0⋅(1−ρ0⋅gP−P0)

Substituting the given values:

ρ=(1.03×103 kg/m3)⋅(1−80.0×1.013×105 Pa−1.03×103 kg/m31.03×103 kg/m3⋅9.81 m/s2)ρ=(1.03×103kg/m3)⋅(1−1.03×103kg/m3⋅9.81m/s280.0×1.013×105Pa−1.03×103kg/m3)

Solving this equation will give us the density of water at the given depth. This method ensures that we account for the increase in pressure with depth, which affects the density of the fluid. It's a great example of how understanding the principles of physics can help us solve real-world problems. If you need further clarification or assistance, feel free to ask!