Thank you for your help. This is one problem with a and b (a) A cook puts 1.70 g of water in a 2.00 L pressure cooker that is then warmed from the kitchen temperature of 20°C to 105°C. What is the pressure (in atm) inside the container?atm(b) What If? If the gauge pressure inside the container cannot exceed 3.00 atm, what is the maximum amount of water (in g) that the cook can put inside the pressure cooker before warming it to105°C?

Answered: A cook puts 1.70 g of water in a 2.00 L…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Thank you for your help. This is one problem with a and b

(a) A cook puts 1.70 g of water in a 2.00 L pressure cooker that is then warmed from the kitchen temperature of 20°C to 105°C. What is the pressure (in atm) inside the container?
atm
(b) What If? If the gauge pressure inside the container cannot exceed 3.00 atm, what is the maximum amount of water (in g) that the cook can put inside the pressure cooker before warming it to
105°C?

Expert Solution

Part (A)

1.7 grams of water will occupy very less volume in comparison with the volume of the container.

So ,initially the cooker will have 1.7 grams of water and atmospheric air.

Initial pressure of air will be same as atmospheric pressure. (101.325 kPa) .

At 105 C , water will be in vapor form.

Since both air and water are in gas from at 105 C , Volume occupied by Water Vapor and Volume occupied by air will be same as volume of the container.

Initial Volume of air and Final Volume of air will be same as volume of the cooker.

$For air, \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} For air initial volume and final volume are same, V_{1} = V_{2} \frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} P_{2} = \frac{P_{1} \times T_{2}}{T_{1}} P_{2} = \frac{101.325 \times (105 + 273)}{(20 + 273)} = 130.72 kPa Partial pressure of air at 105^{0} C, P_{a} = 130.72 Partial Pressure of Water Vapor at 105^{0} C, mass of water m = 1.7 g = 1.7 \times 10^{- 3} kg Volume of water vapor at 105^{0} C = Volume of Cooker = 2 L = 2 \times 10^{- 3} m^{3} PV = mRT P = \frac{mRT}{V} P = \frac{(1.7 \times 10^{- 3}) (\frac{8.314}{18}) (105 + 273)}{2 \times 10^{- 3}} = 148.4 kPa Partial pressure due to water vapor P_{v} = 148.4 kPa Total Pressure P = P_{a} + P_{v} = 130.72 + 148.4 P = 279.12 kPa P = \frac{279.12}{101.325} = 2.75 atm$

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