Explanation Given The volume of helium in the cylinder is 334 mL . The pressure of helium gas is 75 atm . The temperature of helium gas is 23 ° C . Initial mass of helium gas inside the cylinder is 5.225 g . The conversion of ° C into Kelvin is done as, T ° C = T + 273 K Hence, the conversion of 23 ° C into Kelvin is, 23 ° C = 23 + 273 K = 296 K The conversion of mL into L is done as, 1 mL = 10 − 3 L Hence the conversion of 334 mL into L is, 334 mL = 334 × 10 − 3 L =0 .334 L The mass of a gas inside the cylinder can be calculated by using ideal gas equation. P V = n R T P V = ( m M ) R T m = P V M R T Where, P is the pressure of the gas

Chemistry: The Central Science (14th Edition)

14th Edition

ISBN: 9780134414232

Author: Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, Catherine Murphy, Patrick Woodward, Matthew E. Stoltzfus

Publisher: PEARSON

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Chapter 10, Problem 42E

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To determine: The mass of helium that should be released to reduce the pressure of cylinder to 75 atm .

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Chapter 10, Problem 41E

Chapter 10, Problem 43E

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Chemistry: The Central Science (14th Edition)

Ch. 10.2 - Prob. 10.1.1PE Ch. 10.2 - Gallium melts just above room temperature...Ch. 10.2 - Prob. 10.2.1PE Ch. 10.2 - Prob. 10.2.2PE Ch. 10.3 - Prob. 10.3.1PE Ch. 10.3 - Prob. 10.3.2PE Ch. 10.4 - Prob. 10.4.1PE Ch. 10.4 - Prob. 10.4.2PE Ch. 10.4 - Prob. 10.5.1PE Ch. 10.4 - Prob. 10.5.2PE

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