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The rms speed of a sample of gas is increased by \(1 \%\). (a) What is the percent change in the temperature of the gas? (b) What is the percent change in the pressure of the gas, assuming its volume is held constant?

Short Answer

Expert verified
The temperature and pressure both increase by 2.01%.

Step by step solution

01

Understanding the Connection Between RMS Speed and Temperature

The root mean square (rms) speed of gas molecules is given by the formula \( v_{rms} = \sqrt{\frac{3RT}{M}} \), where \( R \) is the gas constant, \( T \) is the temperature, and \( M \) is the molar mass. To find the percent change in temperature, we need to understand that \( v_{rms} \propto \sqrt{T} \).
02

Calculating the Relationship Between RMS Speed and Change in Temperature

If the rms speed increases by 1%, we can represent this change as \( v'_{rms} = 1.01 \times v_{rms} \). Thus, \( \sqrt{T'} = 1.01 \times \sqrt{T} \). Squaring both sides gives \( T' = 1.01^2 \times T \).
03

Calculating Percent Change in Temperature

The percent change in temperature is determined by \( \left( \frac{T' - T}{T} \right) \times 100\% \). Substituting for \( T' \) gives us \( \left( 1.01^2 - 1 \right) \times 100\% = 2.01\% \).
04

Using Ideal Gas Law to Find Percent Change in Pressure

Assuming constant volume and number of moles, we apply the ideal gas law \( PV = nRT \). Since \( V \) and \( n \) are constant, \( P \propto T \). The percent change in pressure is directly equal to the percent change in temperature.
05

Calculating Percent Change in Pressure

Since the temperature increases by 2.01%, the pressure also increases by 2.01% for constant volume.

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Key Concepts

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

RMS Speed
The concept of Root Mean Square (RMS) speed is crucial in understanding the behavior of gas molecules. The RMS speed is defined as the square root of the average of the squares of the individual speeds of gas particles.This quantity provides a useful measure of the speed of gas molecules in a system.To mathematically represent this, the formula is:\[ v_{rms} = \sqrt{\frac{3RT}{M}} \] where:
  • \( v_{rms} \) is the root mean square speed,

  • \( R \) is the gas constant,
  • \( T \) is the temperature in Kelvin,

  • \( M \) is the molar mass of the gas.
A key aspect of RMS speed is its proportionality to the square root of the temperature \( (T) \). This means if the temperature of a gas increases, the RMS speed of the molecules also increases. This relationship allows us to determine changes in temperature when the RMS speed varies, as seen in the exercise, where a 1% increase in the RMS speed results in a 2.01% increase in the temperature.
Temperature and Pressure Relationship
Temperature and pressure share a direct relationship in gases, particularly when the volume is held static.This relationship can be observed from Gay-Lussac's Law, which states that for a constant volume, the pressure of an ideal gas is directly proportional to its temperature. Mathematically, this relationship can be expressed as:\[ P \propto T \]Where:
  • \( P \) is the pressure,

  • \( T \) is the temperature.
If the temperature of a gas inside a fixed volume increases, the pressure increases proportionally. This understanding helps in calculating the change in pressure when the temperature changes within a constant volume system. As derived in the exercise, a 2.01% increase in temperature results in the same percent increase in pressure, demonstrating this direct relationship.
Ideal Gas Law
The Ideal Gas Law is a fundamental principle that combines several gas laws into one comprehensive formula. It connects the four key variables of a gas—pressure, volume, temperature, and the number of moles—into a single equation:\[ PV = nRT \]Where:
  • \( P \) represents the pressure,

  • \( V \) stands for the volume,
  • \( n \) is the amount of substance in moles,

  • \( R \) is the gas constant,
  • \( T \) is the temperature in Kelvin.
This equation assists in predicting how a change in one or more of these variables will affect the others.For constant volume and number of moles, as in this exercise, the Ideal Gas Law simplifies to a direct relationship between pressure and temperature: \( P \propto T \).In scenarios where the RMS speed increases, leading to temperature changes, the Ideal Gas Law aids in explaining concomitant pressure changes.

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Most popular questions from this chapter

An orb weaver spider with a mass of 0.26 g hangs vertically by one of its threads. The thread has a Young's modulus of \(4.7 \times 10^{9} \mathrm{N} / \mathrm{m}^{2}\) and a radius of \(9.8 \times 10^{-6} \mathrm{m}\). (a) What is the fractional increase in the thread's length caused by the spider? (b) Suppose a \(76-\mathrm{kg}\) person hangs vertically from a nylon rope. What radius must the rope have if its fractional increase in length is to be the same as that of the spider's thread?

An ideal gas is kept in a container of constant volume. The pressure of the gas is also kept constant. (a) If the number of molecules in the gas is doubled, does the rms speed increase, decrease, or stay the same? Explain. (b) If the initial rms speed is \(1300 \mathrm{m} / \mathrm{s},\) what is the final rms speed?

If the translational speed of molecules in an ideal gas is doubled, by what factor does the Kelvin temperature change? Explain.

A 155-g aluminum cylinder is removed from a liquid nitrogen bath, where it has been cooled to \(-196^{\circ} \mathrm{C}\). The cylinder is immediately placed in an insulated cup containing 80.0 g of water at \(15.0^{\circ} \mathrm{C}\). What is the equilibrium temperature of this system? If your answer is \(0{ }^{\circ} \mathrm{C}\), determine the amount of water that has frozen. The average specific heat of aluminum over this temperature range is \(653 \mathrm{J} /(\mathrm{kg} \cdot \mathrm{K})\).

A cylindrical flask is fitted with an a irtight piston that is free to slide up and down, as shown in Figure \(17-24\). A mass rests on top of the piston. The initial temperature of the system is \(313 \mathrm{K}\) and the pressure of the gas is held constant at \(137 \mathrm{kPa}\). The temperature is now increased until the height of the piston rises from \(23.4 \mathrm{cm}\) to \(26.0 \mathrm{cm} .\) What is the final temperature of the gas?

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