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Serjik [45]
2 years ago
5

A tank has a volume of 0.1 m3 and is filled with He gas at a pressure of 5 x 106 Pa. A second tank has a volume of 0.15 m' and i

s filled with He gas at a pressure of 6 x 106 Pa. A valve connecting the two tanks is opened. Assumıng He to be a monatomic ideal gas and the walls of the tanks to be adiabatic and rigıd, find the final pressure of the system Hint: Note that the internal energy is constant.
Chemistry
1 answer:
GarryVolchara [31]2 years ago
5 0

Answer:

P=5.6*10^{6} Pa

Explanation:

Consider that, as the system is adiabatic, U_{1}= U_{2} where U1 and U2 are the internal energies before the process and after that respectively.

Consider that: U=H-PV, and that the internal energy of the first state is the sum of the internal energy of each tank.

So, H_{1}-P_{1}V_{1}=H_{2}-P_{2}V_{2}\\H_{1}^{A} -P_{1}^{A}V_{1}^{A}+H_{1}^{B} -P_{1}^{B}V_{1}^{B}=H_{2}-P_{2}V_{2}

Where A y B are the tanks. The enthalpy for an ideal gas is only function of the temperature, as the internal energy is too; so it is possible to assume: H_{1}=H_{2}\\H_{1}^{A}+H_{1}^{B} =H_{2}

So, P_{1}^{A}V_{1}^{A}+P_{1}^{B}V_{1}^{B}=P_{2}V_{2}

Isolating P_{2},

P_{2}=\frac{P_{1}^{A}V_{1}^{A}+P_{1}^{B}V_{1}^{B}}{V_{2}}

V_{2}=V_{1}^{A}+V_{1}^{B}=0.25m^{3}

So,

P_{2}=\frac{5000000*0.1+6000000*0.15}{0.25}=5600000Pa=5.6*10^{6} Pa

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Answer:

The identity does not matter because the variables of Boyle's law do not identify the gas.

Explanation:

The ideal gas law confirms that 22.4 L equals 1 mol.

6 0
2 years ago
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How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos
Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of H_3PO_4 = 54.219 g

Number of atoms of Mg = 7.179\times 10^{23}

Molar mass of H_3PO_4 = 98 g/mol

First we have to calculate the moles of H_3PO_4 and Mg.

\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}

\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol

and,

\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2

From the balanced reaction we conclude that

As, 3 mole of Mg react with 2 mole of H_3PO_4

So, 0.553 moles of Mg react with \frac{2}{3}\times 0.553=0.369 moles of H_3PO_4

From this we conclude that, H_3PO_4 is an excess reagent because the given moles are greater than the required moles and Mg is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of H_2

From the reaction, we conclude that

As, 3 mole of Mg react to give 3 mole of H_2

So, 0.553 mole of Mg react to give 0.553 mole of H_2

Now we have to calculate the volume of H_2  gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies 0.553\times 22.4=12.4L volume of hydrogen gas

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4 0
3 years ago
An oxygen atom has a mass of 2.66*10^-23 and a glass of water has a mass of 0.050 kg. What is the mass of 1 mole of oxygen atoms
Nutka1998 [239]

Answer:

16.0 g; 3.1 mol

Explanation:

(a) Mass of O atoms

Mass = 6.022 × 10^23 atoms × (2.66 × 10^-23 g/1 atom) = 16.0 g

(b) Moles of O atoms

0.050 kg = 50 g

Moles = 50 g × (1 mol/16.0 g) = 3.1 mol

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Answer:

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Explanation:

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