Question

Helium gas at 105.0 kPa and 309.0K is located within a cylinder with a piston. Initially the gas occupies 0.4000 m. While a constant force, F, is applied to the end of the piston so that the pressure inside the cylinder is held constant at 105.0 kPa, 5090.0 J of heat is transferred to the helium gas. The specific enthalpy of helium gas is given by the approximate relation: (kJ/mol) = 0.02087(K) Q=5090.0J Final Conditions P2 = 105.0 kPa Initial Conditions P1 = 105.0 kPa T1 = 309.0K V1 = 0.4000 m2 What is the final temperature of the helium gas? Number T2 =

Answer 1

4 Loco Given Helium gas at iosio kpa and 309.0k is Within a cylinder with a position. Traithella the gas occupies D. 4000 m2 The cylinder is held constant at 1050 kpa, 5090.09. of heat is transferred to the helium gas Ê (kolmol ) = 0.0208 7 (kJ Initially using pu: nht - Where p= pressure VE volume na no. of moles R = Date constant T = Temperature ay 105.0x102x0.40 = n (8.314)(30910) --* 42,000 = n(2569.026)

=> no 42,000 2569.026 In = 16.34 med Qu= ng AT for is obaric process => 5090 =(16.34)( 55 ) (7-309) => 5090 = (16-34) (518314)) (7–309) -> 5090 = 116:34) (20.485) (T-309) => 5090 = 339.6269 (7-309) =7 T-309 = 5090 -Y 5090 = 0 339.6269 => 7-309 = 14.987 => T = 14,987+ 309 T = 323.98 K 2) using pr=nRT. p-pressure ve volume n= no. of mole R= Rate constant 7= Temperature

=Y 105x103 x V = 16. 34 *8.3148 323.98 =Y (105x103) v = 44012.92 V= 44012.92 105x103 V=0.419m? Work done = par = (105x103) 10.419 -0.400 = (105x103) (0.019) - 1995) if work done = 1995 I AU = Q-W = 5090.0 - 1995 = 3095 J. 1 Au= 3095 J