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nasty-shy [4]
2 years ago
14

For a closed system, the change in entropy for a reversible process is _________ the change in entropy for an irreversible proce

ss between the same two states
a. greater than
b. less than
c. the same as
Chemistry
1 answer:
Grace [21]2 years ago
5 0

Answer:

c. the same as

Explanation:

for a closed system:

⇒ ΔS > 0.....irreversible process

⇒ ΔSrev = ∫ dQ/T

for an irreversible process, ΔS is not the same as in a reversible process, since more than one reversible process is needed for the two ΔS to be equal.

⇒ ΔSirrev = ΔSrev1 + ΔSrev2 + ....

but if the initial and final states are the same, ΔS for an irreversible process can be calculated as if it were a reversible process.

⇒ ΔSirrev = ΔSrev

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A 32 L samples of xenon gas at 10°C is expanded to 35 L. Calculate the final temperature.
morpeh [17]

Answer:

              Final Temperature = 36.54 ⁰C

Explanation:

Lets suppose the gas is acting ideally, then according to Charle's Law, "<em>The volume of a fixed mass of gas at constant pressure is directly proportional to the absolute temperature</em>". Mathematically for initial and final states the relation is as follow,

                                                V₁ / T₁  =  V₂ / T₂

Data Given;

                  V₁  =  32 L

                  T₁  =  10 °C = 283.15 K             ∴ K = °C + 273.15

                  V₂  =  35 L

                  T₂  =  ??

Solving equation for T₂,

                         T₂  =  V₂ × T₁  / V₁

Putting values,

                         T₂  =  (35 L × 283.15 K) ÷ 32 L

                         T₂  =  309.69 K     ∴ ( 36.54 °C )

Result:

           As the volume is increased from 32 L to 35 L, therefore, the temperature must have increased from 10 °C to 36.54 °C.

3 0
3 years ago
What does B represent?
Mashutka [201]

Answer: Botanical Acid

Explanation:

8 0
2 years ago
A 80.0 g piece of metal at 88.0°C is placed in 125 g of water at 20.0°C contained in a calorimeter. The metal and water come to
grandymaker [24]

Answer:

The specific heat of the metal is 0.485 J/g°C

Explanation:

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

Mass of the piece of metal = 80.0 grams

Mass of the water = 125 grams

Initial temperature of the metal = 88.0 °C

Initial temperature of water =20.0 °C

Final temperature = 24.7 °C

pecific heat of water is 4.18 J/g*°C

<u>Step 2:</u> Calculate specific heat of the metal

Qgained = -Qlost

Q =m*c*ΔT

Qwater = - Qmetal

m(water)*c(water)*ΔT(water) = -m(metal)*c(metal)*ΔT(metal)

with mass of water = 125 grams

with c( water) = 4.18 J/g°C

with ΔT(water) = T2-T1 = 24.7 - 20 = 4.7°C

with mass of metal = 80.0 grams

with c(metal) = TO BE DETERMINED

with ΔT(metal) = 24.7 - 88.0 = -63.3 °C

125*4.18*4.7 = -80 * C(metal) * -63.3

2455.75 = -80 * C(metal) * -63.3

C(metal) = 2455.75 / (-80*-63.3)

C(metal) = 0.485 J/g°C

The specific heat of the metal is 0.485 J/g°C

3 0
3 years ago
A sample of ice is heated continuously until it becomes a liquid, and then a gas. Its temperature is recorded throughout and a g
padilas [110]

Answer:

The answer to your question is: C. The specific latent heat of fusion

Explanation:

A. The specific latent heat of vaporization  Specific latent heat of vaporization indicates the transition from liquid to vapor, but we are not looking for this definition. This answer is wrong.

B. The specific heat indicates the amount of heat needed to increase the temperature of water 1°C, so this answer is wrong.

C. The specific latent heat of fusion . This heat indicate the transition from solid ie to liquid, so this is the right answer.

D. The internal energy measures the energy of the molecules of a substance, so this answer is wrong.

3 0
2 years ago
Chemistry test bound theories
liq [111]

covalent bond is firmed between two atoms

5 0
2 years ago
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