A gas at 110.0 kPa and 30.0˚C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0˚C and the pressure increased to 440. KPa. What is the new volume?

The answer to your question is   V2 = 0.58 l

Explanation:

Data

Pressure 1 = P1 = 110 kPa                      Pressure 2 = P2 = 440 kPa

Temperature 1 = T1 = 30°C                   Temperature 2 = T2 = 80°C

Volume 1 = V1 = 2 L                               Volume 2 = V2 = ?

Process

- Use the combined gas law to solve this problem

                                P1V1/T1 = P2V2/T2

-Solve for V2

                                V2 = P1V1T2 / T1P2

-Substitution (temperature must be in °K)

                                V2 = (110 x 2 x (80 + 273)) / (30 + 273)(440)

-Simplification

                                V2 = (220 x 353) / (303 x 440)

                                V2 = 77660 / 133320

-Result

                                V2 = 0.58 l

the best answer is d because the strong cohesion is caused by the ability of water molecules to undergo hydrogen bonding.

a is incorrect because when water solidifies, it becomes   ice, which expands and occupies a bigger volume for the same mass of water in its liquid state.

b is incorrect because water has high surface tension. besides bonding to itself it hydrogen bonds to other molecules at the top decreasing possibility for breaking of attractive forces.

c is incorrect because the water molecule has oxygen which is highly electronegative and so pulls the electron pair towards itself creating a dipole moment, making the molecule polar.

e is incorrect because water has strong adhesion, ability to stick to other molecules such as glass because of this molecules having stronger hydrogen bonding than water.