Sure, I can help with that. The formula you provided is for the work done by the system during expansion or compression, where ΔW is the work done, p is the external pressure, and ΔV is the change in volume.

First, we need to convert the pressure from kPa to Pa because the standard unit of pressure in the SI system is Pascal (Pa). 1 kPa = 1000 Pa, so 100.0 kPa = 100,000 Pa.

Next, we need to convert the volume from L to m^3 because the standard unit of volume in the SI system is cubic meter (m^3). 1 L = 0.001 m^3, so 1.50 L = 0.0015 m^3.

Now we can calculate the work done by the system using the formula ΔW = p ΔV:

ΔW = (100,000 Pa) * (0.0015 m^3) = 150 J

The work done by the system is positive because the system is doing work on the surroundings by expanding.

The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. So we can calculate the change in internal energy (ΔU) using the formula ΔU = Q - ΔW, where Q is the heat added to the system:

ΔU = Q - ΔW = 425 J - 150 J = 275 J

So the change in internal energy of the system is 275 J.

Question

Sure, I can help with that. The formula you provided is for the work done by the system during expansion or compression, where ΔW is the work done, p is the external pressure, and ΔV is the change in volume.

First, we need to convert the pressure from kPa to Pa because the standard unit of pressure in the SI system is Pascal (Pa). 1 kPa = 1000 Pa, so 100.0 kPa = 100,000 Pa.

Next, we need to convert the volume from L to m^3 because the standard unit of volume in the SI system is cubic meter (m^3). 1 L = 0.001 m^3, so 1.50 L = 0.0015 m^3.

Now we can calculate the work done by the system using the formula ΔW = p ΔV:

ΔW = (100,000 Pa) * (0.0015 m^3) = 150 J

The work done by the system is positive because the system is doing work on the surroundings by expanding.

The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. So we can calculate the change in internal energy (ΔU) using the formula ΔU = Q - ΔW, where Q is the heat added to the system:

ΔU = Q - ΔW = 425 J - 150 J = 275 J

So the change in internal energy of the system is 275 J.

Knowee AI · Accepted Answer

Sure, I can help with that. The formula you provided is for the work done by the system during expansion or compression, where ΔW is the work done, p is the external pressure, and ΔV is the change in volume.

First, we need to convert the pressure from kPa to Pa because the standard unit of pressure in the SI system is Pascal (Pa). 1 kPa = 1000 Pa, so 100.0 kPa = 100,000 Pa.

Next, we need to convert the volume from L to m^3 because the standard unit of volume in the SI system is cubic meter (m^3). 1 L = 0.001 m^3, so 1.50 L = 0.0015 m^3.

Now we can calculate the work done by the system using the formula ΔW = p ΔV:

ΔW = (100,000 Pa) * (0.0015 m^3) = 150 J

The work done by the system is positive because the system is doing work on the surroundings by expanding.

The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. So we can calculate the change in internal energy (ΔU) using the formula ΔU = Q - ΔW, where Q is the heat added to the system:

ΔU = Q - ΔW = 425 J - 150 J = 275 J

So the change in internal energy of the system is 275 J.

An ideal gas receives 425 J of heat and expands by 1.50 L against an external pressure of 100.0 kPa. What is the change in internal energy of the system? Use-full formula...!ΔW = p ΔV

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