Decreases, stays the same, increases.
The volume decreases because as air is cooled, the individual molecules collectively possess less kinetic energy and the distances between them decrease, thus leading to a decrease in the volume they occupy at a certain pressure (please note that my answer only holds under constant pressure; air, as a gas, doesn't actually have a definite volume).
The mass stays the same because physical processes do not create or destroy matter. The law of conservation of mass is obeyed. You're only cooling the air, not adding more air molecules.
The density decreases because as the volume decreases and mass stays the same, you have the same mass occupying a smaller volume. Density is mass divided by volume, so as mass is held constant and volume decreases, density increases.
Answer:
The value of change in internal energy of the gas = + 1850 J
Explanation:
Work done on the gas (W) = - 1850 J
Negative sign is due to work done on the system.
From the first law we know that Q = Δ U + W ------------- (1)
Where Q = Heat transfer to the gas
Δ U = Change in internal energy of the gas
W = work done on the gas
Since it is adiabatic compression of the gas so heat transfer to the gas is zero.
⇒ Q = 0
So from equation (1)
⇒ Δ U = - W ----------------- (2)
⇒ W = - 1850 J (Given)
⇒ Δ U = - (- 1850)
⇒ Δ U = + 1850 J
This is the value of change in internal energy of the gas.
A) use v=u+at for both
First section, v=27, u=0, a=2.4. You should get 11seconds.
Second section, v=0, u=27, a=-1.3. You should get 21seconds.
This means that the total time is 22seconds.
b) You can either use s=ut+0.5at^2 or v^2=u^2+2as. Personally, I would use the second one as you are not relying on your previous answer.
First section, v=27, u=0, a=2.4. You should get 152m.
Second section, v=0, u=27, a=-1.3. You should get 280m.
This makes your overall displacement 432m.
