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Chapter 16: Problem 5

An ideal gas has a volume of exactly 1 liter at \(1.00 \mathrm{~atm}\) and \(-20^{\circ} \mathrm{C}\). To how many atmospheres of pressure must it be subjected in order to be compressed to \(0.500\) liter when the temperature is \(40^{\circ} \mathrm{C} ?\) $$ \frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}} \quad \text { or } \quad P_{2}=P_{1}\left(\frac{V_{1}}{V_{2}}\right)\left(\frac{T_{2}}{T_{1}}\right) $$ from which $$ P_{2}=(1.00 \mathrm{~atm})\left(\frac{1.00 \mathrm{~L}}{0.500 \mathrm{~L}}\right)\left(\frac{273 \mathrm{~K}+40 \mathrm{~K}}{273 \mathrm{~K}-20 \mathrm{~K}}\right)=2.47 \mathrm{~atm} $$

Short Answer

Expert verified
The gas must be subjected to 2.47 atm of pressure.

Step by step solution

01

- Understand the Combined Gas Law

The combined gas law is expressed as \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \). It shows that the ratio of the product of pressure and volume to the temperature of a gas remains constant if the mass is constant. Our goal is to find \( P_2 \) when \( P_1 \), \( V_1 \), \( T_1 \), \( V_2 \), and \( T_2 \) are given.
02

- Convert Temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin. For \( T_1 = -20^{\circ}C \), we use the formula \( T(K) = T(^{\circ}C) + 273 \). Thus, \( T_1 = -20 + 273 = 253\, \text{K} \). Similarly for \( T_2 = 40^{\circ}C \), \( T_2 = 40 + 273 = 313\, \text{K} \).
03

- Apply the Combined Gas Law

Substitute the known values into the rearranged version of the combined gas law formula to find \( P_2 \): \[ P_2 = P_1 \left(\frac{V_1}{V_2}\right)\left(\frac{T_2}{T_1}\right) \]. Use \( P_1 = 1.00 \text{ atm} \), \( V_1 = 1.00 \text{ L} \), \( V_2 = 0.500 \text{ L} \), \( T_1 = 253 \text{ K} \), and \( T_2 = 313 \text{ K} \).
04

- Compute the Result

Calculate \( P_2 \) using the substituted values: \[ P_2 = 1.00 \times \left(\frac{1.00}{0.500}\right) \times \left(\frac{313}{253}\right) \]. Simplify this step-by-step: \( \left(\frac{1.00}{0.500}\right) = 2.00 \) and \( \left(\frac{313}{253}\right) \approx 1.237 \). Therefore, \( P_2 \approx 2.00 \times 1.237 = 2.47 \text{ atm} \).

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a primary principle that helps us understand how gases behave. It's expressed by the equation \( PV = nRT \). Here, \( P \) represents pressure, \( V \) is volume, \( n \) is the number of moles of gas, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin.
This law assumes gases behave ideally, meaning molecules don't attract or repel each other, and they occupy a negligible volume compared to the container.
  • In reality, no gas is perfectly ideal.
  • However, the Ideal Gas Law is a good approximation for many gases at high temperature and low pressure.
Understanding the Ideal Gas Law provides a foundation for more complex calculations, including those using the Combined Gas Law, which is a derivative tailored for changing conditions of gas pressure, volume, and temperature.
Temperature Conversion
When dealing with gas laws, temperature must always be in Kelvin. The Kelvin scale starts at absolute zero, the coldest possible temperature, making it ideal for calculations involving thermodynamics. To convert from Celsius to Kelvin, simply add 273 to the Celsius temperature. For instance:
  • If a temperature is \(-20^{\circ}C\), then in Kelvin, it's \(-20 + 273 = 253 \, K\).
  • Similarly, if a temperature is \(40^{\circ}C\), in Kelvin, it's \(40 + 273 = 313 \, K\).

This conversion is crucial because the behavior of gases is directly proportional to absolute temperature changes, affecting pressure and volume directly in the Combined Gas Law. Failing to convert temperatures correctly can lead to significant errors in calculations.
Pressure Calculation
Pressure calculation in the context of the Combined Gas Law involves determining the new pressure of a gas when its volume and temperature undergo changes. The formula used here is derived from the law itself: \[ P_2 = P_1 \left(\frac{V_1}{V_2}\right)\left(\frac{T_2}{T_1}\right) \] Where:
  • \( P_1 \) is the initial pressure,
  • \( V_1 \) and \( V_2 \) are the initial and final volumes,
  • \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin.

This calculation helps predict how the gas will behave under new conditions. For example, compressing a gas into half its volume while raising temperature from 253 K to 313 K would result in increasing the pressure to about 2.47 atm. It's a mathematical approach to anticipate how gases react to changes in an enclosed environment.
Volume Compression
Volume compression involves decreasing the space in which a gas is contained, impacting other properties like pressure and temperature. According to the Combined Gas Law, if temperature changes too, these variables are interrelated.
For example, in the given problem, the volume of the gas is decreased from 1.00 L to 0.500 L. Such a change affects the pressure directly, requiring recalculations using our key equation.
  • Reducing the volume puts more molecules in less space.
  • This increases the chances of collisions among molecules and with the container walls, raising pressure if the temperature is constant.

In practice, this principle is vital in various applications like engine cylinders, pressurized canisters, and scientific instruments where controlled pressure is crucial.

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Most popular questions from this chapter

A \(500-\mathrm{mL}\) sealed flask contains nitrogen at a pressure of \(76.00 \mathrm{cmHg} .\) A tiny glass tube lies at the bottom of the flask. Its volume is \(0.50 \mathrm{~mL}\) and it contains hydrogen gas at a pressure of \(4.5 \mathrm{~atm}\). Suppose the glass tube is now broken so that the hydrogen fills the flask. What is the new pressure in the flask?

On a day when atmospheric pressure is \(76 \mathrm{cmHg}\), the pressure gauge on a tank reads the pressure inside to be \(400 \mathrm{cmHg}\). The gas in the tank has a temperature of \(9^{\circ} \mathrm{C}\). If the tank is heated to \(31{ }^{\circ} \mathrm{C}\) by the Sun, and if no gas exits from it, what will the pressure gauge read? $$ \frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}} \quad \text { and } \quad P_{2}=P_{1}\left(\frac{T_{2}}{T_{1}}\right)\left(\frac{V_{1}}{V_{2}}\right) $$ But gauges on tanks usually read the difference in pressure between inside and outside; this is called the gauge pressure. Therefore, $$ P_{1}=76 \mathrm{cmHg}+400 \mathrm{cmHg}=476 \mathrm{cmHg} $$ Also, \(V_{1}=V_{2}\). We then have $$ P_{2}=(476 \mathrm{cmHg})\left(\frac{273+31}{273+9}\right)(1.00)=513 \mathrm{cmHg} $$ The gauge will read \(513 \mathrm{cmHg}-76 \mathrm{cmHg}=437 \mathrm{cmHg}\).

Given \(1000 \mathrm{~mL}\) of helium at \(15^{\circ} \mathrm{C}\) and \(763 \mathrm{mmHg}\), determine its volume at \(-6^{\circ} \mathrm{C}\) and \(420 \mathrm{mmHg}\).

One kilomole of ideal gas occupies \(22.4 \mathrm{~m}^{3}\) at \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm} .(a)\) What pressure is required to compress \(1.00 \mathrm{kmol}\) into a \(5.00 \mathrm{~m}^{3}\) container at \(100^{\circ} \mathrm{C} ?(b)\) If \(1.00 \mathrm{kmol}\) was to be sealed in a \(5.00 \mathrm{~m}^{3}\) tank that could withstand a gauge pressure of only \(3.00 \mathrm{~atm}\), what would be the maximum temperature of the gas if the tank was not to burst?

At \(18^{\circ} \mathrm{C}\) and \(765 \mathrm{mmHg}, 1.29\) liters of an ideal gas has a mass of \(2.71 \mathrm{~g}\). Compute the molecular mass of the gas. Use \(P V=(m / M) R T\) and the fact that \(760 \mathrm{mmHg}=1.00\) atm to obtain $$ M=\frac{m R T}{P V}=\frac{(0.00271 \mathrm{~kg})(8314 \mathrm{~J} / \mathrm{kmol} \cdot \mathrm{K})(291 \mathrm{~K})}{\left[(765 / 760)\left(1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\right]\left(0.00129 \mathrm{~m}^{3}\right)}=50.0 \mathrm{~kg} / \mathrm{kmol} $$
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