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nignag [31]
3 years ago
7

An ideal gas contained in a piston which is compressed. The gas is insulated so that no heat flows into or out of it. 1) What ha

ppens to the temperature of the gas when it is compressed?
Chemistry
1 answer:
Goryan [66]3 years ago
5 0

Answer:Temperature increases

Explanation: As the gas in the container is an ideal gas so it should follow the ideal gas equation, the equation of state.

We know ideal gas equation to be PV=nRT where

   P=pressure

   V=Volume

   T=Temperature

   R=Real gas constant

   n=Number of moles

since the gas is insulated such that no heat goes into or out of the system .

When we compress the ideal gas using a piston, Thermodynamically it means that work is done on the system by the surroundings.

Now as the ideal gas is been compressed so the volume of the gas would decrease and slowly a time will reach when no more gas can be compressed that is there cannot be any further decrease in volume of the gas.

From the equation PV=nRT

Once there is no further compression is possible hence volume becomes constant so pressure of the ideal gas becomes directly proportional to the temperature as n and R are constants. Also as the pressure and volume are inversely related so an decrease in volume would lead to an increase in pressure.

As the ideal  gas is compressed so the pressure of the gas would increase since the gas molecules have smaller volume available after compression hence the gas molecules would quite frequently have collisions with other gas molecules or piston and this collision would lead to increase in speed of the gas molecules and  so the pressure would increase .

The increase in pressure would lead to an increase in temperature as show by the above ideal gas equation because the pressure and temperature are directly related.

So here we can say that work done on the system by surroundings leads to increase in temperature of the system.

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Tems11 [23]

Answer:

If it served you, give me 5 stars please, thank you!

<h3><u>c) 13.29 mL</u></h3>

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2 years ago
Hydrates are formed with _______ compounds
Triss [41]

Answer:

ionic compounds

Explanation:

7 0
3 years ago
How many molecules are in 100 g of C6H120,?*​
Grace [21]

Answer:

3.37 × 10²³ molecules

Explanation:

Given data:

Mass of C₆H₁₂O₆ = 100 g

Number of molecules = ?

Solution:

Number of moles of C₆H₁₂O₆:

Number of moles = mass/molar mass

Number of moles = 100 g/ 180.16 g/mol

Number of moles = 0.56 mol

Number of molecules:

1 mole contain 6.022 × 10²³ molecules

0.56 mol × 6.022 × 10²³ molecules /1 mol

3.37 × 10²³ molecules

6 0
3 years ago
Why does it matter if atoms are divisible ???
Andreyy89

Atoms are divisible contrary to the early beliefs that the smallest "indivisible" matter is an atom. When an atom loses its identity it means that they are divisible. Atoms chemically react with other kinds of atoms thus changing their activity.

They certainly are not that important to our lives, but it’s good to know :)

5 0
3 years ago
A piece of unknown metal with mass 30 g is heated to 110.0 °C and dropped into 100.0 g of water at 20.0 °C. The final temperatur
Ymorist [56]

<u>Answer:</u> The specific heat of metal is 0.821 J/g°C

<u>Explanation:</u>

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]       ......(1)

where,

q = heat absorbed or released

m_1 = mass of metal = 30 g

m_2 = mass of water = 100 g

T_{final} = final temperature = 25°C

T_1 = initial temperature of metal = 110°C

T_2 = initial temperature of water = 20.0°C

c_1 = specific heat of metal = ?

c_2 = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

30\times c_1\times (25-110)=-[100\times 4.186\times (25-20)]

c_1=0.821J/g^oC

Hence, the specific heat of metal is 0.821 J/g°C

8 0
3 years ago
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