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Dovator [93]
3 years ago
11

The water molecules now in your body were once part of a molecular cloud. Only about onemillionth of the mass of a molecular clo

ud is in the form of water molecules, and the mass density of such a cloud is roughly 4.0×10−21 g/cm3 Estimate the volume of a piece of molecular cloud that has the same amount of water as your body.
Physics
1 answer:
Rudik [331]3 years ago
5 0

To solve the problem we first need to define the nomenclature by which there is an X amount of water in our body. Let's call this a variable m, corresponding to the total mass in us and whose units are Kilograms.

Amount_{water} = m (Kg)

We understand that one millionth of this mass, a small fraction of it, is the equivalent of that of a Water Molecule.

Assuming this to have the same body of water in our body that amount should be approximately

m_{water}=10^6

Density of water molecule

\rho = 10^{-18}kg/m^3

V = \frac{m*10^6}{10^{-18}}

V = m*10^{24}m^3

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The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
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Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

We know that

P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v remain same at both side. so we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2 ..............1

let final pressure is P and temp  T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3} ..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3} .............3

divide 2 equation by 3rd equation

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

thus, temperature on left side is 1.48 times the temperature on right

6 0
3 years ago
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