Question

Two moles of nitrogen gas at 25°C, confined within a cylinder by a piston maintaining a constant pressure of 1 atm, is heated with 5.30 kJ of energy. Assume all the energy is used to do work of expansion of the gas at 1 atm. What will be the final temperature of the gas? Recall ∆H = ∆E + P∆V and watch your units!

Answer 1

solution - *- Cose-0 -*-*- initicel Temperature I=85°C I or T = 25+273 | T = 898° Mlumber of moler - 1:37 initial pressure (P) = constenke p=latm → Heat given to the System of constende proasun – Q = AH = 5.30149 I since 1 RJ-100oy = AH = 5:30 X 100 OJ O - AH = 5300) Given - all the heat given to the system use to expansion of the goo. internal energya cone toonde. so

Af=o 7 0 Cose - 7 -&- Assume - Mitrogen gas is ideal gas- and final Temporature of gas is (I). → ideal-gas lover PV=nRT sincer here- 10 = constant na constende (for lose Systemy Racons funk (yos constant i so above equation sacriten as PAV = nRATI OY PC Vg - ~) = OR (T3 - Til | PAV = nR (5-5) 1-0 PAVE NR (Tg-T, → According to first love of thermodynamice AH, AE es quork Related as a 6 AH = 16+ Par

I cuork (u) = par l criven AE=o, & pav=nRl Toloni out in the above equation - AH о+nRAT or AH = RR (Tartig But the valid of AH, 'n, RST to fined Tor Since AH = 53005 n = 2 mole R = 8.314 I мое. К T = 998 K above equation crite as- Torti = ARRA nel 1

To = 29 8kt 55008 ampll x 8:319 $ mol. K T, = 998€ + 5300 K 2x 8.314 To = (298+318.79) K | Tg = 616.74k) To = 616074973 or | To = 343 .799