Find solutions to your problems with the help of IDNLearn.com's knowledgeable users. Our platform provides prompt, accurate answers from experts ready to assist you with any question you may have.
Sagot :
To solve the problem step-by-step, let's break it down clearly. The goal is to analyze the given data and verify the given relationship according to the ideal gas law. The given quantities are:
1. Absolute temperature ([tex]\( T \)[/tex]) = 275 K
2. Pressure ([tex]\( P \)[/tex]) = 105 N/m²
3. Volume ([tex]\( V \)[/tex]) = 0.0225 m³
According to the ideal gas law, the volume [tex]\( V \)[/tex] of a given amount of gas varies directly with its absolute temperature [tex]\( T \)[/tex] and inversely with its pressure [tex]\( P \)[/tex]. The relationship can be expressed with the following formula:
[tex]\[ V \propto \frac{T}{P} \][/tex]
This means:
[tex]\[ V = k \frac{T}{P} \][/tex]
where [tex]\( k \)[/tex] is a proportionality constant. To find [tex]\( k \)[/tex], we'll use the provided values:
Given:
- [tex]\( V = 0.0225 \)[/tex] m³
- [tex]\( T = 275 \)[/tex] K
- [tex]\( P = 105 \)[/tex] N/m²
Using the formula, we can solve for [tex]\( k \)[/tex]:
[tex]\[ 0.0225 = k \frac{275}{105} \][/tex]
Solving the equation for [tex]\( k \)[/tex]:
1. Calculate the ratio [tex]\(\frac{275}{105}\)[/tex]:
[tex]\[ \frac{275}{105} = \frac{55}{21} \][/tex]
2. Substitute the ratio back into the equation:
[tex]\[ 0.0225 = k \cdot \frac{55}{21} \][/tex]
3. Isolate [tex]\( k \)[/tex]:
[tex]\[ k = 0.0225 \cdot \frac{21}{55} \][/tex]
4. Simplify the expression for [tex]\( k \)[/tex]:
[tex]\[ k = 0.0225 \cdot 0.3818 \approx 0.0086 \][/tex]
Therefore, [tex]\( k \approx 0.0086 \)[/tex]. This shows that:
[tex]\[ V = 0.0086 \cdot \frac{T}{P} \][/tex]
Given the proportionate relationship between volume, temperature, and pressure, the previously stated values for the specific gas (0.0225 m³, 275 K, and 105 N/m²) are verified by the ideal gas relationship. The calculated values confirm that if a gas has an absolute temperature of 275 K and pressure of 105 N/m², then the volume would correctly be [tex]\( 0.0225 \)[/tex] m³ according to the ideal gas law.
1. Absolute temperature ([tex]\( T \)[/tex]) = 275 K
2. Pressure ([tex]\( P \)[/tex]) = 105 N/m²
3. Volume ([tex]\( V \)[/tex]) = 0.0225 m³
According to the ideal gas law, the volume [tex]\( V \)[/tex] of a given amount of gas varies directly with its absolute temperature [tex]\( T \)[/tex] and inversely with its pressure [tex]\( P \)[/tex]. The relationship can be expressed with the following formula:
[tex]\[ V \propto \frac{T}{P} \][/tex]
This means:
[tex]\[ V = k \frac{T}{P} \][/tex]
where [tex]\( k \)[/tex] is a proportionality constant. To find [tex]\( k \)[/tex], we'll use the provided values:
Given:
- [tex]\( V = 0.0225 \)[/tex] m³
- [tex]\( T = 275 \)[/tex] K
- [tex]\( P = 105 \)[/tex] N/m²
Using the formula, we can solve for [tex]\( k \)[/tex]:
[tex]\[ 0.0225 = k \frac{275}{105} \][/tex]
Solving the equation for [tex]\( k \)[/tex]:
1. Calculate the ratio [tex]\(\frac{275}{105}\)[/tex]:
[tex]\[ \frac{275}{105} = \frac{55}{21} \][/tex]
2. Substitute the ratio back into the equation:
[tex]\[ 0.0225 = k \cdot \frac{55}{21} \][/tex]
3. Isolate [tex]\( k \)[/tex]:
[tex]\[ k = 0.0225 \cdot \frac{21}{55} \][/tex]
4. Simplify the expression for [tex]\( k \)[/tex]:
[tex]\[ k = 0.0225 \cdot 0.3818 \approx 0.0086 \][/tex]
Therefore, [tex]\( k \approx 0.0086 \)[/tex]. This shows that:
[tex]\[ V = 0.0086 \cdot \frac{T}{P} \][/tex]
Given the proportionate relationship between volume, temperature, and pressure, the previously stated values for the specific gas (0.0225 m³, 275 K, and 105 N/m²) are verified by the ideal gas relationship. The calculated values confirm that if a gas has an absolute temperature of 275 K and pressure of 105 N/m², then the volume would correctly be [tex]\( 0.0225 \)[/tex] m³ according to the ideal gas law.
Your engagement is important to us. Keep sharing your knowledge and experiences. Let's create a learning environment that is both enjoyable and beneficial. For trustworthy answers, rely on IDNLearn.com. Thanks for visiting, and we look forward to assisting you again.