Question

Consider that one mole of hydrogen gas is heated at a constant pressure from 340 K to 480 K. Calculate (a) the energy transferred by heat to the gas, (b) the increase in its internal energy and (c) the work done by the gas. (a) 4.03 kJ; (b) 5.21 kJ; (C) 1.18 kJ (a) 5.03 kJ; (b) 2.86 kJ; (C) 1.18 kJ (a) 4.03 kJ; (b) 2.86 kJ; (c) 1.18 kJ (a) 4.03 kJ; (b) 2.86 kJ; (c) 6.89 kJ

Answer 1

Initial temperature (T) = 340k final temperature (T2) = 480k Constant pressure heat addition process is has & Hydrogen H2 * diatomic gas specific heat ratio for diatomic gas Y=lt. As we know that heat transfer at constant pressure process is given as 9=no&T n=no. of moles (p= r. XR R2 8.314 J 8-1 $ = 1* 1.4 x8-314 (480-340) Universal 1.4-1 gas constant 4.07 Kj ~ 4.03 KJ 9 = 4.03 4.03 kJ ATS } {R2 molik st= change in temperature - Ter Ti b Increase in internal energy du=neudt anGoT ca R 7-1 du= IX 8.314 1.4-1 x (480-340) dus 219 kJ + 2.86 kJ According to first low of thermodynamics I JO = du towa sw a so-du » W = ordu work done Wao-du 4.63-2.86 all7 ? 1:18KI Wsl-18kJ