Question

pressure and gas law quiz
name

1. which gas law addresses the inverse relationship between pressure and volume?
2. which gas law is represented by a syringe that takes in water?
3. which gas law is demonstrated by a balloon expanding as it sits in the sun on a windowsill?
4. what is the formula for pressure? show units in your formula
5. what is the formula to find force? show units for the formula
6. what force is needed on an area of 4m2 to produce a pressure of 16pa?

a. 0.25n
b. 4n
c. 64n
d. 64kg

1. the bottom of a box has an area of 5m2. what is the force of the box to produce a pressure of 20pa?

a. 4n
b. 100n
c. 4kg
d. 100kg

1. box a has a weight of 20n. it is put on top of box b. these two boxes exerts pressure of 60pa. the area box b covers is 3m2. what is the weight of box b?

a. 60n
b. 80n
c. 160n
d. 180n

1. what has to happen to the force exerted if the area doubles but the pressure is unchanged?

a. the force must decrease
b. the force must double
c. the force must increase
d. the force must stay the same

1. the dimensions of a ski are 8cm x 150cm. what is the weight of a skier standing on 2 skis to produce a pressure of 2,500pa?

a 500n
b. 600n
c. 900n
d. 2400n

Explanation:

Step1: Recall the formula for pressure

The formula for pressure is \( P=\frac{F}{A} \), where \( P \) is pressure (in Pascals, \( Pa \)), \( F \) is force (in Newtons, \( N \)), and \( A \) is area (in square meters, \( m^{2} \)).

Step2: Solve for force in question 6

Given \( P = 16Pa \) and \( A=4m^{2} \). Rearranging the formula \( F = P\times A \). Substitute the values: \( F=16\times4 = 64N \).

Step3: Solve for force in question 7

Given \( P = 20Pa \) and \( A = 5m^{2} \). Using \( F=P\times A \), substitute values: \( F=20\times5=100N \).

Step4: Solve for the weight of box B in question 8

First, find the total force \( F \) using \( P = 60Pa \) and \( A = 3m^{2} \). \( F=P\times A=60\times3 = 180N \). Since \( F=F_{A}+F_{B} \) and \( F_{A} = 20N \), then \( F_{B}=F - F_{A}=180 - 20=160N \).

Step5: Analyze the relationship in question 9

From \( F = P\times A \), if \( P \) is constant and \( A \) doubles (\( A_{2}=2A_{1} \)), then \( F_{2}=P\times A_{2}=P\times2A_{1}=2F_{1} \). So the force must double.

Step6: Solve for the weight of the skier in question 10

First, convert the dimensions of one ski to meters: \( 8cm=0.08m \), \( 150cm = 1.5m \). Area of one ski \( A_{1}=0.08\times1.5=0.12m^{2} \). Area of two skis \( A = 2\times0.12 = 0.24m^{2} \). Using \( F=P\times A \) with \( P = 2500Pa \), \( F=2500\times0.24=600N \).

Answer:

1. Boyle's Law
2. Boyle's Law
3. Charles's Law
4. \( P=\frac{F}{A} \) (where \( P \) is in \( Pa \), \( F \) is in \( N \), and \( A \) is in \( m^{2} \))
5. \( F = P\times A \) (where \( F \) is in \( N \), \( P \) is in \( Pa \), and \( A \) is in \( m^{2} \))
6. C. \( 64N \)
7. B. \( 100N \)
8. C. \( 160N \)
9. B. The force must double
10. B. \( 600N \)