Question
1.
A circular piston of radius 0.01 m is compressed by 0.02 meters. What is the change in volume of this compression?
3.14 10-3 m3
6.3 x 10-6 m3
6.3 x 10-4 m3
3.14 x 10-6 m3
2.
If you change the volume of a gas very quickly such that no heat is transferred to the environment, this would be an example of an:
adiabatic process
isovolumetric process
isolated process
isothermal process
3. You blow up a balloon, then squeeze it, and then the balloon pops back out after you finish squeezing it. What are the signs of the work being done on the air in the balloon, in chronological order?
positive, negative, positive
negative, negative, positive
negative, positive, negative
positive, positive, negative
4. .
1000 J of energy is given to a balloon via a heating device. Assuming the pressure is a constant 101,325 Pa, what is the final volume of the balloon, assuming it started at 0.001 m3?
-0.011 m3
-0.001 m3
0.011 m3
0.001 m3
5.
A silly physics student leaves a can of soda in a hot car for a long period of time. When the student returns, the soda can has leaked out into the car. The best description for this situation is:
heat buildup in a hot car
volume build up in an isothermal situation
temperature buildup in an adiabatic situation
pressure buildup in an isovolumetric situation
6.
You slowly crush a full can of juice such that the temperature of the juice does not change, what is changing in this situation? Check all the apply.
Heat
Pressure
Volume
Internal Energy
7. .
A can of gas has a pressure of 101,325 Pa and the volume changes from 0.01 m3 to 0.005 m3. What is the work done?
-1013 J
1,013 J
-507 J
507 J
8. How much pressure is needed to do 1000 J of work when changing a volume of gas by 0.1 m3?
10 Pa
100 Pa
10,000 Pa
101,325 Pa
9. Which of the following definitions of work are accurate? Choose all that apply.
\( W = \frac{F}{A} \)
\( W = P \Delta V \)
\( W = \frac{E}{t} \)
\( W = Fd \)
10. A piston has a volume change of 7 x 10-6 m3. Assuming atmospheric pressure is 101,325 J, what is the work needed to change the piston volume?
0.071 Pa
0.71 Pa
71 Pa
7.1 Pa
A circular piston of radius 0.01 m is compressed by 0.02 meters. What is the change in volume of this compression?
3.14 10-3 m3
6.3 x 10-6 m3
6.3 x 10-4 m3
3.14 x 10-6 m3
2.
If you change the volume of a gas very quickly such that no heat is transferred to the environment, this would be an example of an:
adiabatic process
isovolumetric process
isolated process
isothermal process
3. You blow up a balloon, then squeeze it, and then the balloon pops back out after you finish squeezing it. What are the signs of the work being done on the air in the balloon, in chronological order?
positive, negative, positive
negative, negative, positive
negative, positive, negative
positive, positive, negative
4. .
1000 J of energy is given to a balloon via a heating device. Assuming the pressure is a constant 101,325 Pa, what is the final volume of the balloon, assuming it started at 0.001 m3?
-0.011 m3
-0.001 m3
0.011 m3
0.001 m3
5.
A silly physics student leaves a can of soda in a hot car for a long period of time. When the student returns, the soda can has leaked out into the car. The best description for this situation is:
heat buildup in a hot car
volume build up in an isothermal situation
temperature buildup in an adiabatic situation
pressure buildup in an isovolumetric situation
6.
You slowly crush a full can of juice such that the temperature of the juice does not change, what is changing in this situation? Check all the apply.
Heat
Pressure
Volume
Internal Energy
7. .
A can of gas has a pressure of 101,325 Pa and the volume changes from 0.01 m3 to 0.005 m3. What is the work done?
-1013 J
1,013 J
-507 J
507 J
8. How much pressure is needed to do 1000 J of work when changing a volume of gas by 0.1 m3?
10 Pa
100 Pa
10,000 Pa
101,325 Pa
9. Which of the following definitions of work are accurate? Choose all that apply.
\( W = \frac{F}{A} \)
\( W = P \Delta V \)
\( W = \frac{E}{t} \)
\( W = Fd \)
10. A piston has a volume change of 7 x 10-6 m3. Assuming atmospheric pressure is 101,325 J, what is the work needed to change the piston volume?
0.071 Pa
0.71 Pa
71 Pa
7.1 Pa
Answers
GPT-4o mini
Let's go through each question step by step:
### Question 1:
To find the change in volume due to compression of the circular piston, we'll use the formula for the volume of a cylinder:
\[ V = A \cdot h \]
where \( A \) is the area of the piston, and \( h \) is the compression distance.
1. Calculate the area of the piston \( A \):
\[ A = \pi r^2 = \pi (0.01\, \text{m})^2 = \pi (0.0001\, \text{m}^2) \]
\[ A \approx 3.14 \times 10^{-4}\, \text{m}^2 \]
2. Calculate the change in volume \( \Delta V \):
\[ \Delta V = A \cdot h = 3.14 \times 10^{-4}\, \text{m}^2 \cdot 0.02\, \text{m} \]
\[ \Delta V \approx 6.28 \times 10^{-6}\, \text{m}^3 \]
This round up gives us approximately \( 6.3 \times 10^{-6}\, \text{m}^3 \). Thus, the answer is:
**6.3 x 10^-6 m³**
---
### Question 2:
The process described is one in which the volume of a gas changes but there is no heat exchanged with the surroundings (i.e., the process is fast enough to prevent heat transfer). This defines an:
**adiabatic process**
---
### Question 3:
When you blow up a balloon, you are doing positive work on the air as you increase its volume. When you squeeze it, this is negative work (as the volume decreases). Upon releasing the balloon, it pops back out, and this can be considered as positive work again (the air expands back).
Thus, the signs of the work being done in chronological order are:
**positive, negative, positive**
---
### Question 4:
Using the First Law of Thermodynamics, we can find the final volume \( V_f \):
\[ Q = \Delta U + W \]
At constant pressure \( W = P \Delta V \), where \( \Delta V = V_f - V_i \):
\[ 1000\, J = 0 + 101325\, Pa \cdot (V_f - 0.001\, m^3) \]
We can rearrange it to find \( V_f \):
\[ V_f - 0.001 = \frac{1000}{101325} \]
\[ V_f - 0.001 = 0.009867 \]
\[ V_f = 0.001 + 0.009867 \approx 0.011867\, m^3 \]
This rounds gives us:
**0.011 m³**
---
### Question 5:
The situation refers to pressure buildup inside a can of soda due to heating. When heated, the pressure will build up because of the increase in the kinetic energy of the gas molecules inside the can, but since the volume is not allowed to change significantly, it describes:
**pressure buildup in an isovolumetric situation**
---
### Question 6:
In the process of slowly crushing the can without changing the temperature, heat transfer does not occur (isothermal), but both pressure and volume change:
- Heat: no change
- Pressure: changes
- Volume: changes
- Internal Energy: does not change (for an ideal gas at constant temperature)
So, the answers are:
**Pressure, Volume**
---
### Question 7:
For finding work done during the volume change, we apply the formula:
\[ W = -P \Delta V \]
Where \( \Delta V = V_f - V_i = 0.005\, m^3 - 0.01\, m^3 = -0.005\, m^3 \). Thus:
\[ W = -101325\, Pa \cdot (-0.005\, m^3) = 506.625 J \]
This approximates to:
**507 J**
---
### Question 8:
Using the relationship \( W = P \Delta V \), where \( W = 1000 J \) and \( \Delta V = 0.1 m^3 \), we can rearrange to find pressure:
\[ P = \frac{W}{\Delta V} = \frac{1000\, J}{0.1\, m^3} = 10000\, Pa \]
Thus, the answer is:
**10,000 Pa**
---
### Question 9:
The definitions of work that are accurate:
- \( W = P \Delta V \)
- \( W = Fd \)
So the correct answers are:
**\( W = P \Delta V \) and \( W = Fd \)**
---
### Question 10:
Using the relationship:
\[ W = P \Delta V \]
Where \( \Delta V = 7 \times 10^{-6}\, m^3 \) and \( P = 101325\, Pa \):
\[ W = 101325\, Pa \cdot 7 \times 10^{-6}\, m^3 \]
Calculating gives:
\[ W = 0.709275\, J \approx 0.71 J \]
Thus, in terms of pressure,
**0.71 Pa**. (Note: The unit of work here should actually be in Joules.)
---
Feel free to ask if you have any further questions or need any clarifications!
### Question 1:
To find the change in volume due to compression of the circular piston, we'll use the formula for the volume of a cylinder:
\[ V = A \cdot h \]
where \( A \) is the area of the piston, and \( h \) is the compression distance.
1. Calculate the area of the piston \( A \):
\[ A = \pi r^2 = \pi (0.01\, \text{m})^2 = \pi (0.0001\, \text{m}^2) \]
\[ A \approx 3.14 \times 10^{-4}\, \text{m}^2 \]
2. Calculate the change in volume \( \Delta V \):
\[ \Delta V = A \cdot h = 3.14 \times 10^{-4}\, \text{m}^2 \cdot 0.02\, \text{m} \]
\[ \Delta V \approx 6.28 \times 10^{-6}\, \text{m}^3 \]
This round up gives us approximately \( 6.3 \times 10^{-6}\, \text{m}^3 \). Thus, the answer is:
**6.3 x 10^-6 m³**
---
### Question 2:
The process described is one in which the volume of a gas changes but there is no heat exchanged with the surroundings (i.e., the process is fast enough to prevent heat transfer). This defines an:
**adiabatic process**
---
### Question 3:
When you blow up a balloon, you are doing positive work on the air as you increase its volume. When you squeeze it, this is negative work (as the volume decreases). Upon releasing the balloon, it pops back out, and this can be considered as positive work again (the air expands back).
Thus, the signs of the work being done in chronological order are:
**positive, negative, positive**
---
### Question 4:
Using the First Law of Thermodynamics, we can find the final volume \( V_f \):
\[ Q = \Delta U + W \]
At constant pressure \( W = P \Delta V \), where \( \Delta V = V_f - V_i \):
\[ 1000\, J = 0 + 101325\, Pa \cdot (V_f - 0.001\, m^3) \]
We can rearrange it to find \( V_f \):
\[ V_f - 0.001 = \frac{1000}{101325} \]
\[ V_f - 0.001 = 0.009867 \]
\[ V_f = 0.001 + 0.009867 \approx 0.011867\, m^3 \]
This rounds gives us:
**0.011 m³**
---
### Question 5:
The situation refers to pressure buildup inside a can of soda due to heating. When heated, the pressure will build up because of the increase in the kinetic energy of the gas molecules inside the can, but since the volume is not allowed to change significantly, it describes:
**pressure buildup in an isovolumetric situation**
---
### Question 6:
In the process of slowly crushing the can without changing the temperature, heat transfer does not occur (isothermal), but both pressure and volume change:
- Heat: no change
- Pressure: changes
- Volume: changes
- Internal Energy: does not change (for an ideal gas at constant temperature)
So, the answers are:
**Pressure, Volume**
---
### Question 7:
For finding work done during the volume change, we apply the formula:
\[ W = -P \Delta V \]
Where \( \Delta V = V_f - V_i = 0.005\, m^3 - 0.01\, m^3 = -0.005\, m^3 \). Thus:
\[ W = -101325\, Pa \cdot (-0.005\, m^3) = 506.625 J \]
This approximates to:
**507 J**
---
### Question 8:
Using the relationship \( W = P \Delta V \), where \( W = 1000 J \) and \( \Delta V = 0.1 m^3 \), we can rearrange to find pressure:
\[ P = \frac{W}{\Delta V} = \frac{1000\, J}{0.1\, m^3} = 10000\, Pa \]
Thus, the answer is:
**10,000 Pa**
---
### Question 9:
The definitions of work that are accurate:
- \( W = P \Delta V \)
- \( W = Fd \)
So the correct answers are:
**\( W = P \Delta V \) and \( W = Fd \)**
---
### Question 10:
Using the relationship:
\[ W = P \Delta V \]
Where \( \Delta V = 7 \times 10^{-6}\, m^3 \) and \( P = 101325\, Pa \):
\[ W = 101325\, Pa \cdot 7 \times 10^{-6}\, m^3 \]
Calculating gives:
\[ W = 0.709275\, J \approx 0.71 J \]
Thus, in terms of pressure,
**0.71 Pa**. (Note: The unit of work here should actually be in Joules.)
---
Feel free to ask if you have any further questions or need any clarifications!