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Gas in a container is at a pressure of 1.2 atm and a volume of 2.0 m3.(a) What is the work done on the gas if it expands at constant pressure to twice its initial volume? J(b) What is the work done on the gas if it is compressed at constant pressure to one-quarter of its initial volume? J

Question

Gas in a container is at a pressure of 1.2 atm and a volume of 2.0 m3.(a) What is the work done on the gas if it expands at constant pressure to twice its initial volume? J(b) What is the work done on the gas if it is compressed at constant pressure to one-quarter of its initial volume? J

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Solution

(a) The work done on a gas during a constant pressure process is given by the formula W = PΔV, where P is the pressure and ΔV is the change in volume.

In this case, the gas is expanding to twice its initial volume, so the change in volume is 2.0 m^3 - 1.0 m^3 = 1.0 m^3.

The pressure is given as 1.2 atm, but we need to convert this to pascals (Pa) because the standard unit of work (Joule) is derived from Pa and m^3. 1 atm = 101325 Pa, so 1.2 atm = 1.2 * 101325 = 121590 Pa.

Substituting these values into the formula gives W = 121590 Pa * 1.0 m^3 = 121590 J.

So, the work done on the gas as it expands is 121590 Joules.

(b) If the gas is compressed to one-quarter of its initial volume, the change in volume is 0.5 m^3 - 2.0 m^3 = -1.5 m^3.

The negative sign indicates that the volume is decreasing, which means work is being done on the gas.

Substituting these values into the formula gives W = 121590 Pa * -1.5 m^3 = -182385 J.

So, the work done on the gas as it is compressed is -182385 Joules.

This problem has been solved

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