Question

Documents Pearson Updad Assgrmerti Eimiting. Nos |UNC 217 A gas-filled balloon having a volume of 2.50 L at 1.2 atm and 25°C is allowed to rise to the strato sphere (about 30 km above the surface of Earth), where the temperature and pressure are -23°C and 3.00 x 10 atm, respectively. Calculate the final volume of the balloon 5.35 The temperature of 2.5 L of a gas initially at STP is raised to 250°C at constant volume. Calculate the 5.36 final pressure of the gas in atm. The pressure of 6.0 L of an ideal gas in a flexible container is decreased to one-third of its original pressure, and its absolute temperature is de creased by one-half. What is the final volume of the gas? 5.37 A gas evolved during the fermentation of glucose 5.38 (wine making) has a volume of 0.78 L at 20.1°C and .00 atm. What was the volume of this gas at the fermentation temperature of 36.5°C and 1.00 atm pressure? An ideal gas originally at 0.85 atm and 66°C was allowed to expand until its final volume, presre, and temperature were 94 mL, 0.60 atm, and 45°C, 5.39 respectively. What was its initial volume? 5.40 Calculate its volume (in liters) of 88.4 g of CO at STP .5.41 A gas at 772 mmHg and 35.0 C occupies a volume of 6.85 1 Calculate its volume at STP Dry ice is solid carbon dioxide. A 0.050-g sample of dry ice is placed in an evacuated 4 6-L vessel at 30 C Calculate the pressure inside the vessel after all the dry ice has been comveted to CO gas 5.42

I need help figuring out #5.35, 5.37, 5.45, 5.57, and 5.71

Answer 1

- 5.35 Apply combine gas law ! Pille- Rii 1 We have, P. = 1.2 atr, Vi= 2.50L P2 = 3.00 xlo?atw, T, = 25°C T = -23°c Now, eu love to or V2 = PV, Tz 1 Pz Tz = 1.2 atın x 250L*623°c) - 3.00 x 10 atm x 25°C = 1.2 atun x 250L X (-23+ 27315)K - 3.00015 atm x (25+273.15) K = 1.2 apm x 2.504 * 250:15 - 3.00 x103 att x 298.154 750.45 xio’ 894.45 = 8.39x10² L Thus, final volume of the gas = 18.39x10² L

5. 37 P, = p P2 = P = f T,=T Tz - I - Ni= 60L Apply Pi Na Pa V or V₂ = PV, Tz P, т. P & 6.0L x T/ рхт = Px6.0L X 7x3 8xXx2 = 18.04 = 9.02 Thus, tinal volume of the gas = [9.00]

5.45 we have, P=1.00 atm v=2.10L 5.45 we have, T = 250k p=10x10's atm v = 10L Now, mores of ozone gas is! na Du RT = 10x10 atmx lox 0.0821 y apu melk x 2504 = 4.87*10 mor As 1.00 mal molecule=61022x10 molecules, number of ozone molecules = 4.87x10 mox 6.022x10 moleales/mole = 29.3x10 x 10 molecules = / 2.93x10' molecules / 23 19