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Alex_Xolod [135]
3 years ago
13

Let P and V represent the pressure and volume of the Xe(g) in the container in diagram 3. If a piston is used to reduce the volu

me of the gas to V2 at a constant temperature, what is the new pressure in the container in terms of the original pressure, P ?
Chemistry
1 answer:
Reil [10]3 years ago
7 0

Answer:

2p

Explanation:

To solve this question, we can use Boyle's Law, which states that:

"For a fixed mass of an ideal gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume"

Mathematically:

p\propto \frac{1}{V}

where

p is the pressure of the gas

V is its volume

The equation can be rewritten as

p_1 V_1 = p_2 V_2

where in this problem we have:

p_1 = p is the initial pressure of the Xe(g) gas

V_1=V is the initial volume of the Xe(g) gas

V_2=\frac{V}{2}  is the final volume of the Xe(g) gas

Solving for p2, we find the final pressure of the gas:

p_2=\frac{p_1 V_1}{V_2}=\frac{pV}{V/2}=2p

So, the final pressure is twice the initial pressure.

You might be interested in
A gas contains 75.0 wt% methane, 10.0% ethane, 5.0% ethylene, and the balance water. (a) Calculate the molar composition of this
NeTakaya

Answer:

a)  molar composition of this gas on both a wet and a dry basis are

5.76 moles and 5.20 moles respectively.

Ratio of moles of water to the moles of dry gas =0.108 moles

b) Total air required = 68.51 kmoles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

Explanation:

Let assume we have 100 g of mixture of gas:

Given that :

Mass of methane =75 g

Mass of ethane = 10 g

Mass of ethylene = 5 g

∴ Mass of the balanced water: 100 g - (75 g + 10 g + 5 g)

Their molar composition can be calculated as follows:

Molar mass of methane CH_4}= 16 g/mol

Molar mass of ethane C_2H_6= 30 g/mol

Molar mass of ethylene C_2H_4 = 28 g/mol

Molar mass of water H_2O=18g/mol

number of moles = \frac{mass}{molar mass}

Their molar composition can be calculated as follows:

n_{CH_4}= \frac{75}{16}

n_{CH_4}= 4.69 moles

n_{C_2H_6} = \frac{10}{30}

n_{C_2H_6} = 0.33 moles

n_{C_2H_4} = \frac{5}{28}

n_{C_2H_4} = 0.18 moles

n_{H_2O}= \frac{10}{18}

n_{H_2O}= 0.56 moles

Total moles of gases for wet basis = (4.69 + 0.33 + 0.18 + 0.56) moles

= 5.76 moles

Total moles of gas for dry basis = (5.76 - 0.56)moles

= 5.20 moles

Ratio of moles of water to the moles of dry gas = \frac{n_{H_2O}}{n_{drygas}}

= \frac{0.56}{5.2}

= 0.108 moles

b) If 100 kg/h of this fuel is burned with 30% excess air(combustion); then we have the following equations:

    CH_4 + 2O_2_{(g)} ------> CO_2_{(g)} +2H_2O

4.69         2× 4.69

moles       moles

   C_2H_6+ \frac{7}{2}O_2_{(g)} ------> 2CO_2_{(g)} + 3H_2O

0.33      3.5 × 0.33

moles    moles

    C_2H_4+3O_2_{(g)} ----->2CO_2+2H_2O

0.18           3× 0.18

moles        moles

Mass flow rate = 100 kg/h

Their Molar Flow rate is as follows;

CH_4 = 4.69 k moles/h\\C_2H_6 = 0.33 k moles/h\\C_2H_4=0.18kmoles/h

Total moles of O_2 required = (2 × 4.69) + (3.5 × 0.33) + (3 × 0.18) k moles

= 11.075 k moles.

In 1 mole air = 0.21 moles O_2

Thus, moles of air required = \frac{1}{0.21}*11.075

= 52.7 k mole

30% excess air = 0.3 × 52.7 k moles

= 15.81 k moles

Total air required = (52.7 + 15.81 ) k moles/h

= 68.51 k moles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

5 0
3 years ago
My professor gave me two questions to solve using the Van Der Waals Equation. She told us to solve for P and the second one we h
Fed [463]

Answer:

P=atm

b=\frac{L}{mol}

Explanation:

The problem give you the Van Der Waals equation:

(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT

First we are going to solve for P:

(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}

P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}

Then you should know all the units of each term of the equation, that is:

P=atm

n=mol

R=\frac{L.atm}{mol.K}

a=atm\frac{L^{2}}{mol^{2}}

b=\frac{L}{mol}

T=K

V=L

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

P=\frac{L.atm}{L-L}-atm

Then operate the fraction subtraction:

P=P=\frac{L.atm-L.atm}{L}

P=\frac{L.atm}{L}

And finally you can find the answer:

P=atm

Now solving for b:

(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT

(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}

nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}

b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}

Replacing units:

b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}

Multiplying and dividing units,(please see the second photo below), we have:

b=\frac{L-\frac{L.atm}{atm}}{mol}

b=\frac{L-L}{mol}

b=\frac{L}{mol}

7 0
3 years ago
Which of the following are potential benefits of wind turbines? (may be more than one answer)
Inga [223]
All except for C. And it's good for the environment.
7 0
3 years ago
A cylinder contains 250 g of Helium at 200 K. The external pressure is constant at 1 atm. The temperature of the system is raise
Black_prince [1.1K]

Answer:

There is 96200 J or 96 kJ of heat released.

Explanation:

<u>Step 1: </u>Data given

Mass of helium = 250 grams

Temperature = 200 K

Temperature is raised by 74 K

The heat capacity of helium = 20.8 J/mol*K

Molecular weight for helium = 4 g/mol

<u>Step 2:</u> Calculate moles of Helium

Moles of Helium = mass of Helium / molar mass of helium

Moles of helium = 250 grams / 4g/mol

Moles of helium = 62.5 moles

<u>Step 3:</u> Calculate heat

Q = n*c*ΔT  or

⇒ with Q = the heat released in Joule

⇒ with n = moles of helium = 62.5 moles

⇒ with c = the heat capacity of helium = 20.8 J/mol*K

⇒ with ΔT = the change in temperature = T2 - T1 = 74 K

Q = 62.5 mol * 20.8 J/mol*K * 74 K

Q = 96200 J = 96 kJ

Since the temperature is raised, this is an exothermic reaction. The heat is released.

There is 96200 J or 96 kJ of heat released.

8 0
3 years ago
How many molecules are contained in 125 grams of water, H20?
sergejj [24]
Answer:

18,01528

Explanation:
8 0
2 years ago
Read 2 more answers
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