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skelet666 [1.2K]
3 years ago
7

What is the molar mass of 4.23 g of an elemental gas in a 2.5L container at 282K and 1.4 atm?

Chemistry
1 answer:
balu736 [363]3 years ago
6 0

Answer:

27.98g/mol

Explanation:

Using ideal gas law equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

T = temperature (K)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

According to the information given:

V = 2.5L

P = 1.4 atm

T = 282K

n = ?

Using PV = nRT

n = PV/RT

n = 1.4 × 2.5/0.0821 × 282

n = 3.5/23.1522

n = 0.151mol

Using the formula to calculate molar mass of the elemental gas:

mole = mass/molar mass

Molar mass = mass/mole

Molar mass = 4.23g ÷ 0.151mol

Molar mass = 27.98g/mol

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One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

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Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

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