Question

An ideal gas (which is is a hypothetical gas that conforms to the laws governing gas behavior) confined to a container with a massless piston at the top. (Figure 2) A massless wire is attached to the piston. When an external pressure of 2.00 atm is applied to the wire, the gas compresses from 4.90 to 2.45 L. When the external pressure is increased to 2.50 atm, the gas further compresses from 2.45 to 1.96 L. In a separate experiment with the same initial conditions, a pressure of 2.50 atm was applied to the ideal gas, decreasing its volume from 4.90 to 1.96 L in one step. If the final temperature was the same for both processes, what is the difference between q for the two-step process and for the one-step process in joules? Express your answer with the appropriate units. View Available Hint(s) Value Units Submit

Answer 1

In 1st Process - Two step Process - 1st step, compreession 2nd step, compreession from from 4.901 to 2.45h by external preess. = 2 atm 2.43L to soa6l by ext. preess .= 2.5 atm So work done a - Pext X AV t b change en volume external worth done on preocess system. fore 1st step, sw, = - Qayn x 1.01325 x 103 pu & (2.45-4.90) x abin = 496.5 I [ 13 = 1Pa-m3 ]. for 2nd step, Dwa - 2. Batmx 1.01325 x 105 Pax 11.96-2.45 103 m² x 103 m² = 12.13 Total workAw a Aw, + Ang = 62006] So, According to prest law of thermodynamics AU - AQ + Aw Heat Work done. 10 energy (Au), = (42) + 620.6 I fore 1st Preocess. In and preocess - Strgle step Process - compreession from M.aol to 1.962 by external Preocess Preess .= 2.5atm So, work done, awa - PRAV. = - R. Batm x 1.01325 x 10 Pax (1.96-4.90* x 103 m3 DW = 744.73

Accoreding to 1st Preinciple - (AU),z (4Q), + 744.75 As Initial and final stage aree same, SO DU, ZAU so (AQ), + 62006 = (02)₂ + 744.7] (DQ,)-(AQ)₂ = 12401 ]