Consider the flask apparatus below, which contains 2.00 L of H2 at a pressure of 353 torr and 1.00 L of N2 at an unknown pressure. A flask apparatus is composed of two round flasks connected by a single tube fitted with a closed stopcock. The flask on the left has a volume of 2.00 L and contains H2. The flask on the right has a volume of 1.00 L and contains N2. If the total pressure in the flasks is 373 torr after the stopcock is opened, determine the initial pressure of N2 in the 1.00 L flask. (Assume the final volume is 3.00 L.)

Initial pressure of N2 was 413 torr

Explanation:

Given:

- Initial Pressure of H2 = 353 torr

- Initial Volume of H2 = 2 L

- Final Volume of H2 = 3 L

- Total pressure at end = 373 torr

- Iinitial Volume of N2 = 1 L

- Final Volume of N2 = 3 L

Find:

determine the initial pressure of N2 in the 1.00 L flask.

Solution:

- Lets focus on the H2 at first, from pascal's Law we have:

                                P_1 x V_1 = P_2 x V_2

                                   (353)(2) = (P)(3)

                             P = 353*2 / 3 = 235.333 torr

- Since the total pressure remains constant the Pressure P of the N2 at the end would be

                              373 - 235.33 = 137.667 torr

- Now calculate the original P of N2  , again using Pascal's Law:

                                     (P)(1) = (137.667)(3)

                                   P = 137.667*3 = 413 torr

Initial pressure of N2 was 413 torr

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft