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At $46^{\circ} \mathrm{C}$ a sample of ammonia gas exerts a pressure of 5.3 atm. What is the pressure when the volume of the gas is reduced to one-tenth (0.10) of the original value at the same temperature?

53 atm

Chemistry 101

Chapter 5

Gases

University of Central Florida

Brown University

Lectures

05:03

In physics, a gas is one of the three major states of matter (the others being liquid and solid). A gas is a fluid that does not support tensile stress, meaning that it is compressible. The word gas is a neologism first used by the early 17th-century Flemish chemist J.B. van Helmont, based on the Greek word ("chaos"), the simplest of all the elemental forms of matter.

04:46

In physics, thermodynamics is the science of energy and its transformations. The three laws of thermodynamics state that energy can be exchanged between physical systems as heat and work; that the total energy of a system can be calculated by adding up all forms of energy in the system; that energy spontaneously flows from being localized to becoming dispersed, spread out, or uniform; and that the entropy of an isolated system not in equilibrium will tend to increase over time, approaching a maximum value at equilibrium.

02:50

5.20At 46"C a sam…

01:18

At 53°C a sample of ammoni…

02:30

At $46^{\circ} \mathrm{C}$…

01:12

At $27^{\circ} \mathrm{C}$…

01:25

01:03

A sample of gas at 1.02 at…

A sample of nitrogen gas o…

01:58

A sample of nitrogen gas i…

02:40

A gas sample has a pressur…

01:52

A sample of gas has an ini…

02:01

01:06

Suppose you had a 4.10 - …

01:56

A $10.0 \mathrm{g}$ sample…

All right. So when you did this question, you have to understand this guy he want everyone is equals two p to be to you. So it is given you the 5.3 asked or regional pressure and last take a random number because it didn't say what is the volume of the original. So let's let's take one is easier. And you have to find out the pressure to when it drops to 1 10 So once you do the cross multiplication, you scat. 15. 30. So this problem is pretty easy once you know you have to apply this equation.

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