Grade 11Electromagnetic Induction

Standing waves of frequency 5.0 kHz are produced in a tube filled with oxygen at 300 K. The separation between the consecutive nodes is 3.3 cm. Calculate the specific heat capacities Cp and Cv of the gas.

Simran Bhatia

12 Years agoGrade 11

2 Answers

Kevin Nash

12 Years ago

Sol. f = 5 * 10^3 Hz, T = 300 Hz, λ/2 = 3.3 cm ⇒ λ = 6.6 * 10^-2 m V = fλ = 5 * 10^3 * 6.6 * 10^-2 = (66 * 5) m/s V = fP/f [Pv = nRT ⇒ P = m/mV * Rt ⇒ PM = foRT ⇒ P/fo = RT/m] = √γRT/m(66 * 5) = √γ * 8.3 * 300/32 * 10^-3 ⇒ (66 * 5)^2 = γ * 8.3 * 300/32 * 10^-3 ⇒ γ = (66 * 5)^2 * 32 * 10^-3/8.3 * 300 = 1.3995 C base v= R/γ – 1 = 8.3/0.3995 = 20.7 J/mol-k, C base P = C base V + R = 20.77 + 8.3 = 29.07 J/mol-k.

Navjyot Kalra

12 Years ago

Sol f = 5 * 10^3 Hz, T = 300 Hz, λ/2 = 3.3 cm ⇒ λ = 6.6 * 10^-2 m V = fλ = 5 * 10^3 * 6.6 * 10^-2 = (66 * 5) m/s V = fP/f [Pv = nRT ⇒ P = m/mV * Rt ⇒ PM = foRT ⇒ P/fo = RT/m] = √γRT/m(66 * 5) = √γ * 8.3 * 300/32 * 10^-3 ⇒ (66 * 5)^2 = γ * 8.3 * 300/32 * 10^-3 ⇒ γ = (66 * 5)^2 * 32 * 10^-3/8.3 * 300 = 1.3995 C base v= R/γ – 1 = 8.3/0.3995 = 20.7 J/mol-k, C base P = C base V + R = 20.77 + 8.3 = 29.07 J/mol-k.