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Chemistry: The Central Science | 13th Edition | ISBN: 9780321910417 | Authors: Theodore E. Brown; H. Eugene LeMay; Bruce E. Bursten; Catherine Murphy; Patrick Woodward; Matthew E. Stoltzfus
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You have a gas at 25 C confined to a cylinder with a

Chapter , Problem 10.25

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QUESTION:

You have a gas at 25 C confined to a cylinder with a movable piston. Which of the following actions would double the gas pressure? (a) Lifting up on the piston to double the volume while keeping the temperature constant; (b) Heating the gas so that its temperature rises from 25 C to 50 C, while keeping the volume constant; (c) Pushing down on the piston to halve the volume while keeping the temperature constant.

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QUESTION:

You have a gas at 25 C confined to a cylinder with a movable piston. Which of the following actions would double the gas pressure? (a) Lifting up on the piston to double the volume while keeping the temperature constant; (b) Heating the gas so that its temperature rises from 25 C to 50 C, while keeping the volume constant; (c) Pushing down on the piston to halve the volume while keeping the temperature constant.

ANSWER:

Problem 10.25

You have a gas at 25 C confined to a cylinder with a movable piston. Which of the following actions would double the gas pressure? (a) Lifting up on the piston to double the volume while keeping the temperature constant; (b) Heating the gas so that its temperature rises from 25 C to 50 C, while keeping the volume constant; (c) Pushing down on the piston to halve the volume while keeping the temperature constant.

                                                              Step by Step Solution

Step 1 of 3

(a)

Boyle’s law:

According to this law, the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.

                                             $$ \mathrm{P_{1} V_{1}} = \mathrm{P_{2} V_{2}} $$..................................(1)

                                                 $$ \mathrm{P_{1} } = \text{Initial pressure} $$

                                                 $$ \mathrm{V_{1} } = \text{Initial volume} $$

                                                 $$ \mathrm{P_{2} } = \text{Final pressure} $$

                                                 $$ \mathrm{V_{2} } = \text{Final volume} $$

Rearrange the equation(1) as follows.

                                                  $$ \frac{\mathrm{P_{1} V_{1}}}{\mathrm{V_{2}}}= \mathrm{P_{2}} $$

At constant temperature, the volume of given mass is doubled.

                                                       $$ \mathrm{P}_{2}=\frac{\mathrm{P}_{1} \mathrm{V}_{1}}{2 \mathrm{V}_{1}} $$

Cancel the same constants.

                                                 $$ \mathrm{P}_{2}=\frac{\mathrm{P}_{1} \mathrm{V}_{1}}{2 \mathrm{V}_{1}} $$

                                                  $$ \mathrm{P}_{2}=\frac{1}{2} \mathrm{P}_{1} $$

Therefore, if the volume of the cylinder is doubled at constant temperature then pressure will become half of its initial value. Hence, the option is wrong.

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