Air is being compressed in a cylinder with a volume of 0.025 m^3 under a constant pressure of 3.0*10^5 Pa and the volume of the air in the cylinder is reduced to 0.020 m^3.

a. By how much is the volume of the air reduced?
b. How much work is done in the process?
c. The cylinder is thermally insulated, making the process adaibatic. What is the change in the internal energy of the gas?

I don't know what to do at all! I read my whole textbook about 10 times and can't find anything like this.

a) Subtract the final volume from the initial volume. That is the reduction in volume. Call it delta V

b) Work = P * (delta V)

c) Qin - Wout = change in internal energy (delta U) is a statement of conservation of energy for the gas. Qin is the heat transfered in, which is zero in this case. Wout is the work done BY the gas, which is -W. Use W from part (b)

This means that the answer to (c) is MINUS the answer to (b)

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a) To find the reduction in volume, subtract the final volume from the initial volume:

Reduction in volume (ΔV) = Initial volume - Final volume
ΔV = 0.025 m^3 - 0.020 m^3
ΔV = 0.005 m^3

b) The work done in the process can be determined using the formula:

Work (W) = Pressure (P) * ΔV
W = 3.0 * 10^5 Pa * 0.005 m^3
W = 1500 J

c) Since the process is adiabatic, there is no heat transfer (Qin = 0). The change in internal energy (ΔU) can be calculated using the formula:

ΔU = Qin - Wout
Since Qin is zero, we have:
ΔU = 0 - (-1500 J)
ΔU = 1500 J

Therefore, the change in the internal energy of the gas is 1500 J.

To solve the given problem, we will use the following steps:

a) To find the reduction in volume, we subtract the final volume from the initial volume.

Initial volume = 0.025 m^3
Final volume = 0.020 m^3

Reduction in volume (delta V) = Final volume - Initial volume
= 0.020 m^3 - 0.025 m^3
= -0.005 m^3

Therefore, the volume of the air is reduced by 0.005 m^3.

b) The work done in the process can be calculated using the formula:

Work (W) = Pressure (P) * (delta V)

Given:
Pressure (P) = 3.0 * 10^5 Pa
Reduction in volume (delta V) = -0.005 m^3 (negative sign indicates compression)

Work (W) = (3.0 * 10^5 Pa) * (-0.005 m^3)
= -1500 J

Therefore, the work done in the process is -1500 Joules.

c) Since the cylinder is thermally insulated, the process is adiabatic, meaning there is no heat transfer (Q = 0). According to the conservation of energy for the gas, the change in internal energy (delta U) can be calculated using the equation:

Change in internal energy (delta U) = Heat transfer (Qin) - Work done (Wout)

In this case, Qin = 0, and Wout = -1500 J (from part b). Therefore,

Change in internal energy (delta U) = 0 - (-1500 J)
= 1500 J

Therefore, the change in internal energy of the gas is 1500 Joules.