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sukhopar [10]
2 years ago
11

"What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0"

Chemistry
1 answer:
Artyom0805 [142]2 years ago
4 0

Q: What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0 °C? (Lf = 3.34 x 105 J/kg)

Answer:

-3670.33 J/K

Explanation:

Entropy: This can be defined as the degree of randomness or disorderliness of a substance. The S.I unit of Entropy is J/K.

Mathematically,  change of Entropy can be expressed as,

ΔS = ΔH/T ....................................... Equation 1

Where ΔS = Change of entropy, ΔH = heat change, T = temperature.

ΔH = -(Lf×m).................................... Equation 2

Note: ΔH is negative because heat is lost.

Where Lf = latent heat of ice = 3.34×10⁵ J/kg, m = 3.0 kg, m = mass of water = 3.0 kg

Substitute into equation

ΔH = -(3.34×10⁵×3.0)

ΔH = - 1002000 J.

But T = 0 °C = (0+273) K = 273 K.

Substitute into equation 1

ΔS = -1002000/273

ΔS = -3670.33 J/K

Note: The negative value of ΔS shows that the entropy of water decreases when it is changed to ice at 0 °C

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A 0.529-g sample of gas occupies 125 ml at 60. cm of hg and 25°c. what is the molar mass of the gas?
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<span>Let's </span>assume that the gas has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>

</span><span>Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹) and T is temperature in Kelvin.<span>
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n = ?

<span> R = 8.314 J mol</span>⁻¹ K⁻¹<span>
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<span>
By substitution,
</span></span>79993.4 Pa<span> x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 298 K<span>
                                          n = 4.0359 x 10</span>⁻³ mol

<span>
Hence, moles of the gas</span> = 4.0359 x 10⁻³ mol<span>

Moles = mass / molar mass

</span>Mass of the gas  = 0.529 g 

<span>Molar mass of the gas</span> = mass / number of moles<span>
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<span>                                    = </span>131.07 g mol</span>⁻¹<span>

Hence, the molar mass of the given gas is </span>131.07 g mol⁻¹

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Explanation:

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