Explanation Given Information: Initial Pressure, P 1 = 50 l b f / i n 2 Initial Temperature, T 1 = 200 F = 659.67 R Initial Volume, V 1 = 4 f t 3 Polytropic exponent, n = 1.2 Final temperature, T 2 = 400 F = 859.67 R Obtain the value of the specific heat at constant volume of oxygen. From the Properties of various ideal gases at 25 ° C , 100 k P a , C v = 0.662 k J / k g . K Calculations: Initially, calculate the mass of oxygen in cylinder using following formula, m = P 1 V 1 R T 1 Where, m is mass, P 1 is pressure of side A ,volume is V 1 , and R is gas constant and T 1 is temperature. Substitute, P 1 = 50 l b f / i n 2 , R = 386 f t − l b f / l b m R , V 1 = 48.28 12 l b f . i n / l b m . R , T 1 = 659.67 R m = P 1 V 1 R T 1 m = 50 × 12 3 × 4 48.28 12 × 659.67 m = 0.9042 l b m Therefore, mass of oxygen in cylinder is m = 0.9042 l b m Now calculate the work done during the process 1 − 2 as below, W 1 2 = m R ( T 1 − T 2 ) n − 1 Substitute m = 0.9042 l b m , P 1 = 50 l b f / i n 2 , R = 386 f t − l b f / l b m R , V 1 = 48.28 12 l b f . i n / l b m . R , T 1 = 659.67 R , T 2 = 400 F = 859.67 R , n = 1.2 in equation. W 1 2 = m R ( T 1 − T 2 ) n − 1 W 1 = 0.9042 × 48.28 ( 200 − 400 ) 1.2 − 1 W 1 = − 43654.78 f t − l b f / l b m = − 43654

10th Edition

ISBN: 9781119494966

Author: Borgnakke, C. (claus), Sonntag, Richard Edwin, Author.

Publisher: Wiley,

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Chapter 3, Problem 3.189EP

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Solution Summary: The author explains how to obtain the value of the specific heat at constant volume of oxygen.

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