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A fixed quantity of gas at \(21^{\circ} \mathrm{C}\) exhibits a pressure of 752 torr and occupies a volume of \(5.12 \mathrm{~L}\). (a) Calculate the volume the gas will occupy if the pressure is increased to \(1.88 \mathrm{~atm}\) while the temperature is held constant. (b) Calculate the volume the gas will occupy if the temperature is increased to \(175^{\circ} \mathrm{C}\) while the pressure is held constant.

Short Answer

Expert verified
(a) The volume of the gas will be \(2.6985\ L\) when the pressure is increased to \(1.88\ atm\) while the temperature is held constant. (b) The volume of the gas will be \(7.8127\ L\) when the temperature is increased to \(175^{\circ} C\) while the pressure is held constant.

Step by step solution

01

Convert the given temperature into Kelvin

To convert the given temperature (in Celsius) into Kelvin, we can use the formula: \(T(K) = T(^\circ C) + 273.15\) For the initial temperature of \(21^\circ C\) : \(T(K)= 21 + 273.15 = 294.15 K\)
02

Convert the given pressure into a consistent unit

Convert the given pressure from torr to atm to maintain consistency with the pressure value that will be increased. We can use the following conversion factor: \(1\ atm = 760\ torr\) Initial pressure in atm: \(P_1 = \frac{752 \ torr}{760} = 0.9895\ atm\) Increased pressure: \(P_2 = 1.88\ atm\)
03

Use Boyle's Law to solve for the volume (Part a)

Boyle's Law states that under constant temperature, the volume of a gas is inversely proportional to its pressure. The formula for Boyle's Law is: \(P_1V_1 = P_2V_2\) We are given \(P_1\), \(V_1\), and \(P_2\), and we are asked to calculate the volume \(V_2\) when the pressure is increased while the temperature is held constant. We can rearrange the formula of Boyle's Law to find \(V_2\): \(V_2 = \frac{P_1V_1}{P_2}\) Plugging in the given values: \(V_2 = \frac{(0.9895\ atm)(5.12\ L)}{1.88\ atm} = 2.6985\ L\) So, the volume of the gas will be \(2.6985\ L\) when the pressure is increased to \(1.88\ atm\) while the temperature is held constant.
04

Use Charles' Law to solve for the volume (Part b)

Charles' Law states that under constant pressure, the volume of a gas is directly proportional to its temperature. The formula for Charles' Law is: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) We are given the initial volume, initial temperature, and final temperature, and we are asked to calculate the volume \(V_2\) when the temperature is increased while the pressure is held constant. First, we need to convert the final temperature to Kelvin: \(T_2 = 175^\circ C + 273.15 = 448.15\ K\) Next, we can rearrange the formula of Charles' Law to find \(V_2\): \(V_2 = V_1 \cdot \frac{T_2}{T_1}\) Plugging in the given values: \(V_2 = 5.12\ L \cdot \frac{448.15\ K}{294.15\ K} = 7.8127\ L\) So, the volume of the gas will be \(7.8127\ L\) when the temperature is increased to \(175^\circ C\) while the pressure is held constant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boyle's Law
Understanding Boyle's Law is crucial when dealing with the compression and expansion of gases. This law describes how the volume of gas tends to decrease as pressure increases when temperature remains constant. Robert Boyle, a pioneer of modern chemistry, made this discovery in the 17th century. Essentially, Boyle's Law is mathematically stated as:
\(P_1V_1 = P_2V_2\),
where \(P_1\) and \(V_1\) are the initial pressure and volume, and \(P_2\) and \(V_2\) are the new pressure and volume after the change.
When you're tasked with calculating how volume changes with pressure, remember that they are inversely proportional. In other words, if you double the pressure, the volume will halve, provided the system's temperature is kept constant. This simple yet powerful relationship has significant applications in our daily lives, from the operation of syringes to the functionality of our lungs when we breathe.
Charles' Law
Charles' Law is all about the relationship between the volume of gas and its temperature. This law tells us that if we keep the pressure of a gas constant, its volume will change in direct proportion to its temperature.
Named after French scientist Jacques Charles, Charles' Law is expressed through the formula:
\(\frac{V_1}{T_1} = \frac{V_2}{T_2}\).
This means if you heat a balloon, it expands because the volume of the gas inside it increases with the increase in temperature. Remember, it's crucial to measure temperature on an absolute scale (in Kelvin) because the volume of a gas becomes zero at absolute zero. Charles' Law is not just academic; it's the principle behind hot air balloons rising and falling and why car tires can become overinflated in summer or appear to lose pressure in winter.
Temperature Conversion
Converting temperature between Celsius and Kelvin is a simple yet essential process in understanding gas laws. Kelvin is the unit of absolute temperature and is one of the fundamental units in physics. The conversion formula is:
\(T(K) = T(^\fC) + 273.15\).
This equation suggests that 0 degrees Celsius corresponds to 273.15 Kelvin, which is the basis of the Kelvin scale. Since many gas law formulas require temperature in Kelvin, this conversion is the first step in numerous calculations. If the exercise gives you a temperature in Celsius (like \(21^\fC\)), you convert it into Kelvin before using it in calculations requiring temperature, ensuring accuracy in describing the behavior of a gas under various conditions.
Pressure-Volume Relationship
The pressure-volume relationship is a cornerstone concept in understanding the physical behavior of gases. According to Boyle's Law, for a given mass of an ideal gas at constant temperature, the volume and pressure of the gas are inversely proportional. This means as pressure increases, volume decreases and vice versa. In mathematical terms, if volume goes up by a factor (for example, doubles), then pressure goes down by the same factor (halves), given temperature remains unchanged.

This relationship is essential for various applications, including calculating the dosage in medical syringe injections or understanding the mechanics of piston engines. The ability to predict how changing pressure will affect volume enables engineers and scientists to design better systems and machines that utilize the physics of gases.

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Most popular questions from this chapter

(a) Amonton's law expresses the relationship between pressure and temperature. Use Charles's law and Boyle's law to derive the proportionality relationship between \(P\) and \(T\). (b) If a car tire is filled to a pressure of \(32.0 \mathrm{lbs} / \mathrm{in}^{2}\) (psi) measured at \(75^{\circ} \mathrm{F}\), what will be the tire pressure if the tires heat up to \(120^{\circ} \mathrm{F}\) during driving?

Calculate each of the following quantities for an ideal gas: (a) the volume of the gas, in liters, if \(1.50 \mathrm{~mol}\) has a pressure of \(1.25 \mathrm{~atm}\) at a temperature of \(-6^{\circ} \mathrm{C}\) (b) the absolute temperature of the gas at which \(3.33 \times 10^{-3} \mathrm{~mol}\) occupies \(478 \mathrm{~mL}\) at 750 torr; (c) the pressure, in atmospheres, if \(0.00245 \mathrm{~mol}\) occupies \(413 \mathrm{~mL}\) at \(138^{\circ} \mathrm{C}\); (d) the quantity of gas, in moles, if \(126.5 \mathrm{~L}\) at \(54^{\circ} \mathrm{C}\) has a pressure of \(11.25 \mathrm{kPa}\)

(a) Place the following gases in order of increasing average molecular speed at \(25^{\circ} \mathrm{C}: \mathrm{Ne}, \mathrm{HBr}, \mathrm{SO}_{2}, \mathrm{NF}_{3}, \mathrm{CO}\). (b) Calculate the rms speed of \(\mathrm{NF}_{3}\) molecules at \(25^{\circ} \mathrm{C}\). (c) Calculate the most probable speed of an ozone molecule in the stratosphere, where the temperature is \(270 \mathrm{~K}\).

The molar mass of a volatile substance was determined by the Dumas-bulb method described in Exercise 10.53. The unknown vapor had a mass of \(0.846 \mathrm{~g}\); the volume of the bulb was \(354 \mathrm{~cm}^{3}\), pressure 752 torr, and temperature \(100^{\circ} \mathrm{C}\). Calculate the molar mass of the unknown vapor.

Nitrogen and hydrogen gases react to form ammonia gas as follows: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ At a certain temperature and pressure, \(1.2 \mathrm{~L}\) of \(\mathrm{N}_{2}\) reacts with \(3.6 \mathrm{~L}_{2}\) of \(\mathrm{H}_{2}\). If all the \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) are consumed, what volume of \(\mathrm{NH}_{3}\), at the same temperature and pressure, wi

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