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Chapter 3: Problem 110

\(3.110\) A balloon filled with helium, initially at \(27^{\circ} \mathrm{C}, 1\) bar, is released and rises in the atmosphere until the helium is at \(17^{\circ} \mathrm{C}, 0.9\) bar. Determine, as a percent, the change in volume of the helium from its initial volume.

Short Answer

Expert verified
The volume of helium increases by 7.4%.

Step by step solution

01

Understand the Ideal Gas Law

The Ideal Gas Law states that for a given amount of gas, the relationship between pressure (P), volume (V), and temperature (T) is given by: \[ PV = nRT \] where n is the number of moles, and R is the universal gas constant.
02

Convert Temperatures to Kelvin

Temperatures must be in Kelvin to use the Ideal Gas Law. Convert the initial and final temperatures: \[ T_1 = 27^{\circ}C + 273.15 = 300.15 \ K \] \[ T_2 = 17^{\circ}C + 273.15 = 290.15 \ K \]
03

Apply the Ideal Gas Law Initially

Using the Initial conditions and the Ideal Gas law, express the initial volume: \[ P_1 V_1 = nRT_1 \] \[ V_1 = \frac{nRT_1}{P_1} \]
04

Apply the Ideal Gas Law at Final State

Using the Final conditions and the Ideal Gas Law, express the final volume: \[ P_2 V_2 = nRT_2 \] \[ V_2 = \frac{nRT_2}{P_2} \]
05

Express Volume Ratio

Combine initial and final volume equations to find the ratio of final volume to initial volume: \[ \frac{V_2}{V_1} = \frac{nRT_2/P_2}{nRT_1/P_1} = \frac{T_2 P_1}{T_1 P_2} \]
06

Substitute Values in Volume Ratio

Substitute the given values for temperatures and pressures: \[ \frac{V_2}{V_1} = \frac{290.15 \ K \times 1 \ bar}{300.15 \ K \times 0.9 \ bar} = \frac{290.15}{270.135} = 1.074 \]
07

Determine Percent Change in Volume

The percent change in volume can be determined by: \[ \%\Delta V = \left( \frac{V_2 - V_1}{V_1} \right) \times 100 = \left(1.074 - 1\right) \times 100 = 7.4\% \]

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Key Concepts

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

temperature conversion to Kelvin
To apply the Ideal Gas Law, temperatures must be converted to Kelvin.
The Kelvin scale starts at absolute zero, the point where all molecular motion stops.
Converting Celsius temperatures to Kelvin ensures a positive and consistent scale.
Simply add 273.15 to the Celsius temperature to convert it:
\[ T(K) = T(掳C) + 273.15 \]
For example, if the temperature is 27掳C:
\[ T(1) = 27 + 273.15 = 300.15 \]
Similarly, for 17掳C:
\[ T(2) = 17 + 273.15 = 290.15 \]
Now, both temperatures are in Kelvin, ready to be used in the Ideal Gas Law calculations.
volume ratio
Using the Ideal Gas Law, we can relate the initial and final states of the gas.
From the law: \[ PV = nRT \]
We first solve for volume in terms of pressure, temperature, and constants:
\[ V(1) = \frac{nRT(1)}{P(1)} \]
\[ V(2) = \frac{nRT(2)}{P(2)} \]
To find how volume changes, we use the Volume Ratio:
\[ \frac{V(2)}{V(1)} = \frac{T(2)P(1)}{T(1)P(2)} \]
Plug in the given values:
\[ \frac{V(2)}{V(1)} = \frac{290.15 \, K \times 1 \, bar}{300.15 \, K \times 0.9 \, bar} = 1.074 \]
This ratio helps us see by how much the volume changes.
percent change calculation
To express the change in volume as a percentage, we use the volume ratio we calculated.
We found:
\[ \frac{V(2)}{V(1)} = 1.074 \]
The percent change in volume can be found by:
\[ \% \Delta V = \left( \frac{V(2) - V(1)}{V(1)} \right) \times 100 \]
Simplifying this:
\[ \% \Delta V = \left( 1.074 - 1 \right) \times 100 = 7.4\% \]
A positive value indicates an increase, so the volume of the helium increased by 7.4%.

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Most popular questions from this chapter

A piston-cylinder assembly contains water, initially saturated liquid at \(150^{\circ} \mathrm{C}\). The water is heated at constant temperature to saturated vapor. (a) If the rate of heat transfer to the water is \(2.28 \mathrm{~kW}\), determine the rate at which work is done by the water on the piston, in \(\mathrm{kW}\). (b) If in addition to the heat transfer rate given in part (a) the total mass of water is \(0.1 \mathrm{~kg}\), determine the time, in \(\mathrm{s}\), required to execute the process.

Ammonia in a piston-cylinder assembly undergoes two processes in series. Initially, the ammonia is saturated vapor at \(p_{1}=100 \mathrm{lbf} / \mathrm{in} .^{2}\) Process \(1-2\) involves cooling at constant pressure until \(x_{2}=75 \%\). The second process, from state 2 to state 3 , involves heating at constant volume until \(x_{3}=100 \%\). Kinetic and potential energy effects are negligible. For \(1.2 \mathrm{lb}\) of ammonia, determine (a) the heat transfer and work for Process 1-2 and (b) the heat transfer for Process 2-3, all in Btu.

Air contained in a piston-cylinder assembly, initially at 2 bar, \(200 \mathrm{~K}\), and a volume of \(1 \mathrm{~L}\), undergoes a process to a final state where the pressure is 8 bar and the volume is \(2 \mathrm{~L}\). During the process, the pressure-volume relationship is linear. Assuming the ideal gas model for the air, determine the work and heat transfer, each in kJ.

A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at \(27^{\circ} \mathrm{C}\) and a gage pressure of \(300 \mathrm{kPa}\). The gas is heated, and the gage pressure at the final state is \(367 \mathrm{kPa}\). Determine the final temperature, in \({ }^{\circ} \mathrm{C}\). The local atmospheric pressure is \(1 \mathrm{~atm}\).

Two kg of oxygen fills the cylinder of a piston-cylinder assembly. The initial volume and pressure are \(2 \mathrm{~m}^{3}\) and 1 bar, respectively. Heat transfer to the oxygen occurs at constant pressure until the volume is doubled. Determine the heat transfer for the process, in kJ, assuming the specific heat ratio is constant, \(k=1.35\). Kinetic and potential energy effects can be ignored.
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