Question

Calculate the temperature of an argon sample at 55.4 kPa and 18.6 L if it occupies 25,8 L at 75.0°C and 411 kPa. Multiple Choice 95.0°C 85.1°C 77.2°C 72.9°C < Prey 9 of 10 !!! Nah >

Answer 1

Page No Solution from given Ideal gas equation we solve problem! the We Know that PV=ART where V n pressure in kpa of volume occupied by Gas in term of O ltre. Number of moles of gas Gos constant - 8.3145 mole-t R - T Temperature in Kelvin K. To'c=273k Given - at Volume = 25.8L Temperature = 75.0°c = (75 +273)K = 348k pressure = 41.1kpa R= 8.3145 L-kpa mole - Kelvin First we have to find out the moles of argon gas, PV = nRT na pv RT

na 41.1 kp a x 25.8L 8.3145 Lepa x 348 Kelvin mole-Kelvin n= 41.1 X 25.B 3.3145 X34B mole D=0.366476512B mole n=0.3 665 mole We have to gind out the Temperature when volume = 18.6L pressure = 55.4 kpa Pv=ART Te pv DR 55.4 kpa x 18.6L 0.3665 mole X Q 3145 kcpa-L mole- Kelvin Kelvin T = 55.4 x 18.6 0.3665 XB-3145 = 33 8.15 Kelui's 7 273 K2 oc) TE/338.15–273 c T= 65.15°C T= 65.2°C Abs.