The complete question is this: This figure (Figure 1) shows a container that is sealed at the top by a movable piston. Inside the container is an ideal gas at 1.00 atm, 20.0 ∘C, and 1.00 L. This information will apply to the first three parts of this problem.
A) What will the pressure inside the container become if the piston is moved to the 1.20 L mark while the temperature of the gas is kept constant?
Explanation:
It is given that,
= 1 atm, = 1 L
= ? , = 1.20 L
As the temperature is constant. Hence, find the value of as follows.
=
=
= 0.833 atm
Thus, we can conclude that the pressure inside the container is 0.833 atm.
Answer:
66.475kJ
Explanation:
Work done is the product of the force and perpendicular distance in direction of the force.
Work done = Force × distance
Given
Weight = 700lb
Distance = 70ft
1lb force = 4.448N
700lb force = 700(4.448) = 3113.6N
Distance = 70(0.305) = 21.35m
Work done = 3113.6×21.35
Work done ≈ 66,475Joules
Work done = 66.475kJ
Answer:
49 N (d)
Explanation:
w= mg = 5 kg * 9.8 m/s^2 = 49 N