1. 1.67 moles of an ideal monatomic gas(C m,p. = 5R/2) at 26.50 o C is expanded

Question

1. 1.67 moles of an ideal monatomic gas(Cm,p. = 5R/2) at 26.50oC is expanded at aconstant pressure of 745 Torr until its final temperature is78.33oC. Calculate, in kJ, for the system: a) w, b) q,c) DU, d) DH
Please help. Have a couple of questions of this type.Would like to see how this one is solved so that I can solve theothers. Thanks alot. 1. 1.67 moles of an ideal monatomic gas(Cm,p. = 5R/2) at 26.50oC is expanded at aconstant pressure of 745 Torr until its final temperature is78.33oC. Calculate, in kJ, for the system: a) w, b) q,c) DU, d) DH
Please help. Have a couple of questions of this type.Would like to see how this one is solved so that I can solve theothers. Thanks alot.

Explanation / Answer

Given no. of moles , n = 1.67 moles Pressure , P = 745 torr                   = 745 * 133.3 Pa        Since 1 torr = 133.3Pa                   = 99308.5 Pa Initial temperature , T = 26.5 o C = 26.5 + 273 = 299.5K Final Temperature , T' = 78.33 o C = 78.33+273 = 351.33K Work done , W = n R dT                          =1.67mol * 8.314Js * ( 351.33-299.5 )                         = 719.62 J                          =0.71962 KJ q = nC p dT    = 1.67 * (5R / 2 ) * 51.83    = 1.67 * 2.5 * 8.314 * 51.83     = 1799.068 J     = 1.799KJ U = q - W       = 1.799 KJ - 0.71962KJ       = 1.07938 KJ     = 1.799KJ U = q - W       = 1.799 KJ - 0.71962KJ       = 1.07938 KJ H = U + nRdT        = 1.07938 KJ + 1.67 *8.314 * 51.83 * 10 ^-3        = 1.799 KJ

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