A) Calculate the ratio (pure number) of the kinetic energy to the potential ener

Question

A) Calculate the ratio (pure number) of the kinetic energy to the potential energy of a simple harmonic oscillator when its displacement is half its amplitude.

B) A system consists of 3.2 mol of the monatomic gas neon (which may be treated as an ideal gas). Initially the system has a temperature of 290K and a volume of 1.2 cubic meters. heat is added to this system at constant pressure until the volume triples, then more heat is added at constant volume until the pressure doubles. Find the total heat added to the system.

Explanation / Answer

(A). The ratio of the kinetic energy to the potential energy of a simple harmonic oscillator is           K / U = [( 1/ 2)m 2 ( A 2 - x 2 ) ] / [ ( 1/ 2) m 2 x 2 ]                    = ( A 2 - x 2 ) / x 2 Given x = A / 2 Subsitute this value we get ,K / U = [3A 2 / 4 ] / [A 2 / 4]                                                     = 3 i.e., required ratio is ,K : U = 3 : 1 (B).Number of moles n = 3.2 mol Initial volume V = 1.2 m 3 Initial temperature T = 290 K At constant pressure :- Final volume V ' = 3V Heat added Q = n x Cp x ( T ' - T ) Where C p = Specific heat at constant pressure                    = 2.5 R                    = 2.5 x 8.314                    = 20.785 J / mol K               T ' = Final tempearture                    = ( V ' / V ) T                    = 3 x 290                    = 870 K Substitute the values we get Q = 38576.96 J At constant volume :- Final pressure P " = 3P Heat added Q ' = n x C v x ( T" - T ' ) Where C v = Specific heat at constant volume                    = 1.5 R                    = 1.5 x 8.314                    = 12.471 J / mol K              T " = Final tempearture                    = ( P " / P ) T '                    = 3 x 870                    = 2610 K Substitute the values we get Q ' = 69438.5 J Required heat = Q + Q '                           = 108.015 x 10 3 J Given x = A / 2 Subsitute this value we get ,K / U = [3A 2 / 4 ] / [A 2 / 4]                                                     = 3 i.e., required ratio is ,K : U = 3 : 1 (B).Number of moles n = 3.2 mol Initial volume V = 1.2 m 3 Initial temperature T = 290 K At constant pressure :- Final volume V ' = 3V Heat added Q = n x Cp x ( T ' - T ) Where C p = Specific heat at constant pressure                    = 2.5 R                    = 2.5 x 8.314                    = 20.785 J / mol K               T ' = Final tempearture                    = ( V ' / V ) T                    = 3 x 290                    = 870 K Substitute the values we get Q = 38576.96 J At constant volume :- Final pressure P " = 3P Heat added Q ' = n x C v x ( T" - T ' ) Where C v = Specific heat at constant volume                    = 1.5 R                    = 1.5 x 8.314                    = 12.471 J / mol K              T " = Final tempearture                    = ( P " / P ) T '                    = 3 x 870                    = 2610 K Substitute the values we get Q ' = 69438.5 J Required heat = Q + Q '                           = 108.015 x 10 3 J Where C v = Specific heat at constant volume                    = 1.5 R                    = 1.5 x 8.314                    = 12.471 J / mol K              T " = Final tempearture                    = ( P " / P ) T '                    = 3 x 870                    = 2610 K Substitute the values we get Q ' = 69438.5 J Required heat = Q + Q '                           = 108.015 x 10 3 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.