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Chapter 18: Problem 13

If a certain amount of ideal gas occupies a volume \(V\) at STP on earth, what would be its volume (in terms of \(V )\) on Venus, where the temperature is \(1003^{\circ} \mathrm{C}\) and the pressure is 92 atm?

Short Answer

Expert verified
The gas volume on Venus would be approximately \(\frac{V}{4.675}\).

Step by step solution

01

Understand the Problem and Given Conditions

We need to find the volume of an ideal gas on Venus in terms of its original volume at Standard Temperature and Pressure (STP) on Earth. At STP, temperature is 273.15 K and pressure is 1 atm. On Venus, the temperature is 1003掳C, which is equivalent to 1276.15 K, and the pressure is 92 atm.
02

Use Ideal Gas Law to Relate Conditions

The Ideal Gas Law is given by \(PV = nRT\). Because the amount of gas and the gas constant \(R\) are constant, we can set up a ratio between the initial (Earth) and final (Venus) conditions: \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\). We will solve for the unknown \(V_2\).
03

Substitute Known Values into the Equation

Use the initial conditions \(P_1 = 1\) atm, \(V_1 = V\), \(T_1 = 273.15\) K, and final conditions \(P_2 = 92\) atm, \(T_2 = 1276.15\) K. Substitute these into the formula: \(\frac{1 \cdot V}{273.15} = \frac{92 \cdot V_2}{1276.15}\).
04

Solve for the Final Volume \(V_2\)

Rearrange the equation to solve for \(V_2\): \(V_2 = \frac{V \cdot 273.15 \cdot 92}{1276.15}\). Calculate \(V_2\) using these values to express it in terms of \(V\): \(V_2 = \frac{V \cdot 92}{4.675}\), simplifying this gives \(V_2 \approx \frac{V}{4.675}\).

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Key Concepts

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

Standard Temperature and Pressure (STP)
Understanding Standard Temperature and Pressure, or STP, is crucial for solving gas law problems. STP is a set of agreed-upon conditions used to make comparisons between different sets of measurements.
In chemistry, STP is defined as a temperature of 273.15 K (0掳C) and a pressure of 1 atm. These conditions are used as a reference point for the behavior of gases, known as ideal gases.
The behavior of these gases under STP can be predicted using the Ideal Gas Law, which enables us to calculate volumes, pressures, or temperatures by knowing at least one variable.
Temperature and Pressure on Venus
The environmental conditions on Venus are drastically different from those on Earth.
The temperature on Venus can soar up to 1003掳C, which is converted to Kelvin as 1276.15 K. This extremely high temperature is due to the thick, greenhouse gas-rich atmosphere trapping heat.
Furthermore, the pressure on Venus is about 92 atm, much higher than Earth's atmospheric pressure. Such extreme conditions affect the behavior of gases significantly.
  • The high temperature increases the energy of gas molecules, causing them to move more vigorously.
  • The high pressure compresses gases into much smaller volumes compared to more moderate conditions like those on Earth.
Volume Comparison of Gases
Comparing the volumes of gases under different conditions involves applying the Ideal Gas Law: \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature.
When dealing with volume comparisons, we often use the ratio form of the Ideal Gas Law: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \). This helps determine how a gas鈥檚 volume changes when moved from one environment to another.
For instance, when calculating the volume of a gas on Venus in terms of STP volume from Earth, the conditions on Venus (high pressure and temperature) must be plugged into the equation.
  • At STP: \( P_1 = 1 \) atm, \( T_1 = 273.15 \) K.
  • On Venus: \( P_2 = 92 \) atm, \( T_2 = 1276.15 \) K.
By simplifying and rearranging the equation, we find that the volume on Venus \( V_2 \) becomes approximately \( \frac{V}{4.675} \), indicating a significantly reduced volume due to the extreme Venusian conditions.

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Most popular questions from this chapter

How Close Together Are Gas Molecules? Consider an ideal gas at \(27^{\circ} \mathrm{C}\) and 1.00 atm pressure. To get some idea how close these molecules are to each other, on the average, imagine them to be uniformly spaced, with each molecule at the center of a small cube. (a) What is the length of an edge of each cube if adjacent cubes touch but do not overlap? (b) How does this distance compare with the diameter of a typical molecule? (c) How does their separation compare with the spacing of atoms in solids, which typically are about 0.3 \(\mathrm{nm}\) apart?

It is possible to make crystalline solids that are only one layer of atoms thick. Such "two-dimensional" crystals can be created by depositing atoms on a very flat surface. (a) If the atoms in such a two-dimensional crystal can move only within the plane of the crystal, what will be its molar heat capacity near room temperature? Give your answer as a multiple of \(R\) and in \(\mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\) ) At very low temperatures, will the molar heat capacity of a two-dimensional crystal be greater than, less than, or equal to the result you found in part (a)? Explain why.

How many moles are in a 1.00 -kg bottle of water? How many molecules? The molar mass of water is 18.0 \(\mathrm{g} / \mathrm{mol}\) .

You have several identical balloons. You experimentally determine that a balloon will break if its volume exceeds 0.900 L. The pressure of the gas inside the balloon equals air pressure 1.00 atm). (a) If the air inside the balloon is at a constant temperature of \(22.0^{\circ} \mathrm{C}\) and behaves as an ideal gas, what mass of air can you blow into one of the balloons before it bursts? (b) Repeat part (a) if the gas is helium rather than air.

An automobile tire has a volume of 0.0150 \(\mathrm{m}^{3}\) on a cold day when the temperature of the air in the tire is \(5.0^{\circ} \mathrm{C}\) and atmospheric pressure is 1.02 atm. Under these conditions the gauge pressure is measured to be 1.70 atm (about 25 \(\mathrm{lb} / \mathrm{in.} .\) ). After the car is driven on the highway for 30 min, the temperature of the air in the tires has risen to \(45.0^{\circ} \mathrm{C}\) and the volume has risen to 0.0159 \(\mathrm{m}^{3}\) . What then is the gauge pressure?
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