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Reil [10]
3 years ago
13

Calculate the approximate volume of a 0.6000mol sample of gas at 288.15K and a pressure of 1.10atm.

Chemistry
1 answer:
11111nata11111 [884]3 years ago
6 0

Answer:

The volume of the sample of the gas is found to be 12.90 L.

Explanation:

Given pressure of the gas = P = 1.10 atm

Number of moles of gas = n = 0.6000 mole

Temperature = T = 288.15 K

Assuming the volume of the gas to be V liters

The ideal gas equation is shown below

\textrm{PV} =\textrm{nRT} \\1.10 \textrm{ atm}\times V \textrm{ L} = 0.6000 \textrm{ mole}\times 0.0821 \textrm{ L.atm.mol}^{-1}.K^{-1}\times 288.5\textrm{K} \\\textrm{V} = 12.90 \textrm{ L}

Volume occupied by gas = 12.90 L

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<h3>How to determine the volume occupied by the gas in a balloon </h3>

Let suppose that <em>flammable</em> hydrogen behaves ideally. GIven the molar mass (M), in kilograms per kilomole, and mass of the gas (m), in kilograms. The volume occupied by the gas (V), in cubic centimeters, is found by the equation of state for <em>ideal</em> gases:

V = \frac{m\cdot R_{u}\cdot T}{P\cdot M}   (1)

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If we know that m = 0.239\,kg, R_{u} = 8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K}, T = 273.15\,K, P = 101.325\,kPa and M = 2.02\,\frac{kg}{kmol}, then the volume of the balloon is:

V = \frac{(0.239\,kg)\cdot \left(8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K} \right)\cdot (273.15\,K)}{\left(101.325\,kPa\right)\cdot \left(2.02\,\frac{kg}{kmol} \right)}

V = 2.652\,m^{3} (2652\,L)

The volume of the balloon is approximately 2652 liters. \blacksquare

To learn more on ideal gases, we kindly invite to check this verified question: brainly.com/question/8711877

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