a rigid tank whose volume is unknown is divided into two parts by a partition. THERMODYNAMICS HELP!!!!?
A rigid tank of unknown volume is divided into two parts by a partition. One side of the tank contains an ideal gas at a temperature of 927 degrees Celsius, while the other side is evacuated and has a volume twice that of the part containing the gas. Once the partition is removed, the gas expands to fill the entire tank. Afterward, heat is applied to the gas until its pressure equals the initial pressure. Can you please help me determine the final temperature of the gas? I'm unsure how to approach this problem.
2 Answers
Ti = 927 deg. C = 1200 K
Vf = 3*Vi
Pi = P
Pf = P
Using gas ideal equation,
PiVi/Ti = PfVf/Tf, Pi = Pf
Vf/Ti = Vf/Tf
Tf = Ti*(Vf/Vi)
Tf = 1200*(3)
Tf =3600 K = 3327 deg. C
i dont comprehend if u are acquainted with 2d grade equations so im providing a a lot less complicated answer merely divide 1512 into factors: 1512=2x2x2x3x3x3x7 now, u ought to multiply a number of them and something else as a thanks to get the sum seventy 8 social gathering; attempt 2x2x2x3=24 and the last 3x3x7=sixty 3 and u dont get seventy 8 besides, after some tries u will locate 2x2x3x3=36 and the last 2x3x7=40 2 which have the sum seventy 8 so the nrs are 40 2 and 36 this problem is unquestionably solved with 2d grade equations" enable the first nr be x and the 2d be seventy 8-x u have x*(seventy 8-x)=1512 78x-x^2-1512=0 or x^2-78x+1512=0. then u keep on with the formulation, if u comprehend it
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