/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 Compute the volume of \(8.0\) g ... [FREE SOLUTION] | 91Ó°ÊÓ

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Compute the volume of \(8.0\) g of helium \((M=4.0 \mathrm{~kg} / \mathrm{kmol})\) at 15 \({ }^{\circ} \mathrm{C}\) and \(480 \mathrm{mmHg}\). Use \(P V=(m / M) R T\) to obtain $$ V=\frac{m R T}{M P}=\frac{(0.0080 \mathrm{~kg})(8314 \mathrm{~J} / \mathrm{kmol} \cdot \mathrm{K})(288 \mathrm{~K})}{(4.0 \mathrm{~kg} / \mathrm{kmol})\left[(480 / 760)\left(1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\right]}=0.075 \mathrm{~m}^{3}=75 \text { liters } $$

Short Answer

Expert verified
The volume of helium is 75 liters.

Step by step solution

01

Understanding the Equation

The problem uses the equation \( P V = \frac{m}{M} R T \) to calculate the volume \( V \) of helium gas. Here, \( P \) is pressure, \( V \) is volume, \( m \) is mass, \( M \) is molar mass, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.
02

Identifying Known Values

We are given: mass \( m = 8.0 \text{ g} = 0.0080 \text{ kg}\), molar mass \( M = 4.0 \text{ kg/mol} \), temperature \( T = 15^\circ C = 288 \text{ K} \), and pressure \( P = 480 \text{ mmHg} \). We need to convert the pressure to \( ext{N/m}^2 \) by using \( 1 ext{ atm} = 760 \text{ mmHg} = 1.01 \times 10^5 \text{ N/m}^2 \).
03

Convert Pressure to Proper Units

Calculate \( P \) in \( ext{N/m}^2 \):\[ P = \left(\frac{480}{760}\right) \times 1.01 \times 10^5 \text{ N/m}^2 \].
04

Plug Values into the Volume Equation

Utilize the rearranged ideal gas law equation for volume, \( V = \frac{(m \cdot R \cdot T)}{(M \cdot P)} \). Substitute the known values into this formula: \[ V = \frac{(0.0080 \text{ kg})(8314 \text{ J/Kmol} \cdot \text{K})(288 \text{ K})}{(4.0 \text{ kg/Kmol}) \cdot (\frac{480}{760} \cdot 1.01 \times 10^5 \text{ N/m}^2)} \].
05

Calculate the Volume

Perform the calculations step-by-step: \[ V = \frac{0.0080 \times 8314 \times 288}{4.0 \times (\frac{480}{760} \cdot 1.01 \times 10^5)} = 0.075 \text{ m}^3 \].
06

Convert Volume to Liters

Convert the volume from cubic meters to liters by multiplying by 1000 (as 1 m³ = 1000 liters): \( 0.075 \text{ m}^3 \times 1000 = 75 \text{ liters} \).

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Key Concepts

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

Helium gas calculation
Understanding how to calculate the volume of helium gas using the ideal gas law can be straightforward once you grasp each component. The formula utilized here is derived from the famous Ideal Gas Law, expressed as \( PV = \frac{m}{M} RT \). In this context,
  • \( P \) stands for pressure
  • \( V \) represents volume
  • \( m \) is the mass of the gas
  • \( M \) refers to the molar mass of helium
  • \( R \) is the universal gas constant
  • \( T \) signifies temperature measured in Kelvin
For this specific case, you recalibrate the equation to solve explicitly for volume \( V \). This becomes \( V = \frac{(m \times R \times T)}{(M \times P)} \). By plugging in the known values—such as mass, molar mass, pressure, and temperature—you can compute the volume of helium. It’s essential to keep the units consistent to attain an accurate outcome. The ultimate calculation leads to the volume expressed in cubic meters or liters.
Molar mass
Molar mass is a measure expressing the weight of one mole of a given substance, typically in kilograms per kmol or grams per mol. For helium, a noble gas characterized by its lightweight, the molar mass is given as \( 4.0 \text{ kg/kmol} \). This is a crucial value in deriving calculations involving gases as it allows conversion from mass (either in grams or kilograms) to moles, using the relation:
\[m = M \cdot n\]Here,
  • \( m \) is mass,
  • \( M \) functions as the molar mass, and
  • \( n \) represents the number of moles.
Understanding molar mass facilitates linking between measurable mass and chemical quantities in your calculations, making it indispensable in the field of chemistry. Given this molar mass, you can translate between macroscopic mass and the number of molecules participating in a gas reaction or under set conditions.
Gas constant
The universal gas constant, often symbolized as \( R \), stands at the heart of the Ideal Gas Law. The given value is \( 8314 \text{ J/kmol} \cdot \text{K} \), which remains consistent across all ideal gas calculations globally. This constant bridges various physical properties of gases—linking pressure, volume, and temperature to substance mass and molar mass through its specific energy units.
Incorporating \( R \) into your calculations, as seen in the helium gas calculation, ensures that all energy-related aspects of gases align within the equation. This standardized constant makes it possible for students to apply it universally, whether computing gas volumes or exploring thermodynamic processes in chemistry and physics. Always maintain the proper unit balance when using \( R \) to prevent miscalculations.
Pressure conversion
Pressure conversion is essential to align different units into a standard form suitable for equations like the Ideal Gas Law. Atmospheric pressure is commonly recorded in millimeters of mercury (mmHg) or in units of atmospheres (atm); however, scientific calculations often employ newtons per meter squared (\(\text{N/m}^2\)).
To convert pressure from mmHg to \(\text{N/m}^2\), use the conversion factor for atmospheric pressure: \(1 \text{ atm} = 760 \text{ mmHg} = 1.01 \times 10^5 \text{ N/m}^2\).
  • Apply the fraction: \(\frac{480}{760}\) to transition from mmHg to atm.
  • Multiply by \(1.01 \times 10^5\) to finalize the pressure in \(\text{N/m}^2\).
This conversion is crucial to maintain consistency with the units of the gas constant and the Ideal Gas Law equation, ensuring accurate results across all calculations. Proper understanding of unit conversions further assists in preventing errors and enhancing conceptual clarity pertaining to pressure in scientific inquiries.

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Most popular questions from this chapter

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 $$ P_{2}=(476 \mathrm{~cm} \mathrm{Hg})\left(\frac{273+31}{273+9}\right)(1.00)=513 \mathrm{cmHg} $$ The gauge will read \(513 \mathrm{cmHg}-76 \mathrm{cmHg}=437 \mathrm{cmHg}\).

A certain mass of hydrogen gas occupies \(370 \mathrm{~mL}\) at \(16^{\circ} \mathrm{C}\) and 150 kPa. Find its volume at \(-21^{\circ} \mathrm{C}\) and \(420 \mathrm{kPa}\). $$ \begin{array}{c} \frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}} \quad \text { leads to } \quad V_{2}=V_{1}\left(\frac{P_{1}}{P_{2}}\right)\left(\frac{T_{2}}{T_{1}}\right) \\\ V_{2}=(370 \mathrm{~mL})\left(\frac{150 \mathrm{kPa}}{420 \mathrm{kPa}}\right)\left(\frac{273 \mathrm{~K}-21 \mathrm{~K}}{273 \mathrm{~K}+16 \mathrm{~K}}\right)=115 \mathrm{~mL} \end{array} $$

A \(5000-\mathrm{cm}^{3}\) tank contains an ideal gas \((M=40 \mathrm{~kg} / \mathrm{kmol})\) at a gauge pressure of \(530 \mathrm{kPa}\) and a temperature of \(25^{\circ} \mathrm{C}\). Assuming atmospheric pressure to be \(100 \mathrm{kPa}\), what mass of gas is in the tank?

A 2.0-mg droplet of liquid nitrogen is present in a \(30 \mathrm{~mL}\) tube as it is sealed off at very low temperature. What will be the nitrogen pressure in the tube when it is warmed to \(20{ }^{\circ} \mathrm{C}\) ? Express your answer in atmospheres. (M for nitrogen is \(28 \mathrm{~kg} / \mathrm{kmol}\).) Use \(P V=(m / M) R T\) to find $$ \begin{aligned} P &=\frac{m R T}{M V}=\frac{\left(2.0 \times 10^{-6} \mathrm{~kg}\right)(8314 \mathrm{~J} / \mathrm{kmol} \cdot \mathrm{K})(293 \mathrm{~K})}{(28 \mathrm{~kg} / \mathrm{kmol})\left(30 \times 10^{-6} \mathrm{~m}^{3}\right)}=5800 \mathrm{~N} / \mathrm{m}^{2} \\ &=\left(5800 \mathrm{~N} / \mathrm{m}^{2}\right)\left(\frac{1.0 \mathrm{~atm}}{1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}}\right)=0.057 \mathrm{~atm} \end{aligned} $$

Imagine a cylinder containing \(0.500 \mathrm{~m}^{3}\) of gas sealed in with a movable piston. If the gas in the cylinder is heated so that its temperature goes from \(250 \mathrm{~K}\) to \(500 \mathrm{~K}\) keeping the pressure constant, if the volume changes, what will its new value be? [Hint: Use Charles' Law.]

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