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A cylinder contains 0.100 mol of an ideal monatomic gas.

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 36E Chapter 19

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 36E

A cylinder contains 0.100 mol of an ideal monatomic gas. Initially the gas is at 1.00 X 105 Pa and occupies a volume of 2.50 X 10-3 m3. (a) Find the initial temperature of the gas in kelvins. (b) If the gas is allowed to expand to twice the initial volume, find the final temperature (in kelvins) a

Step-by-Step Solution:

Solution 36E Step 1: The given gas is an ideal gas, hence must obey the governing equations of ideal gas. We know that, PV = nRT --------------(1) Where, P is the pressure, V is the volume, N is the amount of the substance of gas, R is the gas constant, T is the temperature. Here the values of P, V, n, R, T are 1 × 10 Pa, 2.50 × 10 3 m , 0.100 mol, 1 1 8.314 J K mol respectively. Step 2: From equation (1), 5 3 1 × 10 × 2.50 × 10 = 0.100 × 8.314 × T T = (1 × 10 × 2.50 × 10 ) / (0.100 × 8.314) T = 2.50 × 10 / 0.8314 = 300.69 K The initial temperature T we found as 300.69 K. 1 Step 3: Now the gas expands to twice it’s initial volume. If the expansion is isothermal, which means the initial and final temperatures are same, then, From another equation of ideal gas, P V P V 1 1= 2 2 ----------------(2) T1 T 2 P V = P V 1 1 2 2 P V = P × 2V 1 1 2 1 1 × 10 × 2.50 × 10 3 = P × 2 × 2.50 × 10 3 2 P = (2.50 × 10 ) / (4.50 × 10 ) = 0.55555 × 10 Pa. 5 2

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Chapter 19, Problem 36E is Solved
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Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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A cylinder contains 0.100 mol of an ideal monatomic gas.

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