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\(\bullet$$\bullet\) If a certain amount of ideal gas occupies a volume \(V\) at \(S T P\) 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 volume on Venus is approximately 0.05075V.

Step by step solution

01

Understand STP Conditions

STP (Standard Temperature and Pressure) conditions imply a temperature of 0°C (273.15 K) and a pressure of 1 atm. This means our initial volume, \( V \), is when the gas is under these conditions.
02

Use the Ideal Gas Law

The ideal gas law, \( PV = nRT \), can be rearranged to compare two states of the same gas. So we have \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \), where \( V_1 = V \) (at STP), \( T_1 = 273.15 \, \text{K} \), \( P_1 = 1 \, \text{atm} \), \( P_2 = 92 \, \text{atm} \), and \( T_2 = 1003^{\circ}\text{C} = 1276.15\, \text{K} \).
03

Solve for Final Volume

Rearrange the equation \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \) to find \( V_2 \), which is the volume on Venus: \[ V_2 = \frac{P_1V_1T_2}{P_2T_1} = \frac{1 \cdot V \cdot 1276.15}{92 \cdot 273.15} \approx \frac{1276.15V}{25139.8}. \] Performing this calculation gives approximately \( V_2 \approx 0.05075V \).

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Key Concepts

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

Standard Temperature and Pressure
Standard Temperature and Pressure (STP) are the conditions universally adopted for comparing gas volumes. At STP, the temperature is 0°C, which translates to 273.15 Kelvin, and the pressure is 1 atmosphere (atm). These conditions make it easy to calculate and predict the behavior of gases since many of their properties rely on temperature and pressure.
Using STP in calculations is essential when applying the Ideal Gas Law, as it provides a consistent basis from which changes can be measured. Consider STP as a baseline to explore how gases behave under different conditions.
Volume Calculation
Calculating the volume of a gas under varying conditions often involves the Ideal Gas Law, expressed as \( PV = nRT \). However, when comparing two states of a gas, another useful form is derived: \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \). This rearrangement helps solve problems like finding the new volume of a gas when its temperature and pressure change.
To use this formula effectively, identify each variable:
  • Initial Pressure \( P_1 \) and Temperature \( T_1 \) are from STP conditions.
  • Pressures and temperatures at the final state are \( P_2 \) and \( T_2 \).

Plug these values into the equation to find the new volume \( V_2 \). This approach is powerful, as it accounts for work done on or by the gas during the transformation.
Pressure and Temperature
Pressure and temperature are critical in determining gas volume due to their direct impact on molecular motion and spacing. In the Ideal Gas Law, pressure \( P \) and temperature \( T \) are directly proportional to the volume \( V \), provided the number of gas molecules and constant \( R \) remain unchanged.
When pressure increases and temperature decreases, gases are compressed, decreasing their volume. Conversely, if temperature rises while pressure is constant, gases expand to larger volumes.
  • Pressure is often measured in atmospheres (atm).
  • Temperature is usually expressed in Kelvin (K) for calculations.

For the exercise, with pressure soaring to 92 atm and temperature surging to 1276.15 K on Venus, the gas volume drastically changes, showcasing how sensitive gases are to both these properties.

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Most popular questions from this chapter

\(\bullet\) \(\bullet\) A diver observes a bubble of air rising from the bottom of a lake (where the absolute pressure is 3.50 atm ) to the surface (where the pressure is 1.00 atm). The temperature at the bottom is \(4.0^{\circ} \mathrm{C}\) and the temperature at the surface is \(23.0^{\circ} \mathrm{C}\) . $$\begin{array}{l}{\text { (a) What is the ratio of the volume of the bubble as it reaches }} \\ {\text { the surface to its volume at the bottom? (b) Would it be safe }} \\ {\text { for the diver to hold his breath while ascending from the bottom of the lake to the surface? Why or why not? }}\end{array}$$

\(\bullet\) A cylinder contains 0.250 mol of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) gas at a temperature of \(27.0^{\circ} \mathrm{C}\) . The cylinder is provided with a frictionless piston, which maintains a constant pressure of 1.00 atm on the gas. The gas is heated until its temperature increases to \(127.0^{\circ} \mathrm{C}\) . Assume that the \(\mathrm{CO}_{2}\) may be treated as an ideal gas. (a) Draw a \(p V\) diagram of this process. (b) How much work is done by the gas in the process? (c) On what is this work done? (d) What is the change in internal energy of the gas? (e) How much heat was supplied to the gas? (f) How much work would have been done if the pressure had been 0.50 atm?

\(\bullet\) Helium gas with a volume of 2.60 \(\mathrm{L}\) under a pressure of 1.30 atm and at a temperature of \(41.0^{\circ} \mathrm{C}\) is warmed until both the pressure and volume of the gare are doubled. (a) What is the final temperature? (b) How many grams of helium are there? The molar mass of helium is 4.00 \(\mathrm{g} / \mathrm{mol}.\)

\(\bullet\) Lung volume. The total lung volume for a typical person is 6.00 L. A person fills her lungs with air at an absolute pressure of 1.00 atm. Then, holding her breath, she compresses her chest cavity, decreasing her lung volume to 5.70 L. What is the pressure of the air in her compressed lungs, assuming that the temperature of the air remains constant?

\(\bullet$$\bullet\) A cylinder 1.00 \(\mathrm{m}\) tall with inside diameter 0.120 \(\mathrm{m}\) is used to hold propane gas (molar mass 44.1 \(\mathrm{g} / \mathrm{mol} )\) for use in a barbecue. It is initially filled with gas until the gauge pressure is \(1.30 \times 10^{6} \mathrm{Pa}\) and the temperature is \(22.0^{\circ} \mathrm{C} .\) The temperature of the gas remains constant as it is partially emptied out of the tank, untill the gauge pressure is \(2.50 \times 10^{5}\) Pa. Calculate the mass of propane that has been used.

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