(a) The work done on the gas can be calculated using the formula:

W = PΔV

where:
W is the work done,
P is the pressure, and
ΔV is the change in volume.

The pressure P is given as 0.800 atm, but we need to convert this to pascals (Pa) because the SI unit for work (Joule) is derived from pascals and liters. We know that 1 atm = 101325 Pa, so:

P = 0.800 atm * 101325 Pa/atm = 81060 Pa

The change in volume ΔV is the final volume minus the initial volume:

ΔV = 3.00 L - 6.00 L = -3.00 L

We also need to convert this to cubic meters (m^3) because the SI unit for volume is m^3. We know that 1 L = 0.001 m^3, so:

ΔV = -3.00 L * 0.001 m^3/L = -0.003 m^3

Now we can calculate the work done:

W = PΔV = 81060 Pa * -0.003 m^3 = -243.18 J

So, the work done on the gas is -243.18 J.

(b) The change in internal energy ΔU of the gas can be calculated using the first law of thermodynamics:

ΔU = Q - W

where:
ΔU is the change in internal energy,
Q is the heat transferred to the system, and
W is the work done by the system.

In this case, the heat Q is given as -360 J (the negative sign indicates that energy is leaving the gas), and we've calculated the work W as -243.18 J. So:

ΔU = Q - W = -360 J - (-243.18 J) = -116.82 J

So, the change in internal energy of the gas is -116.82 J.

Question

(a) The work done on the gas can be calculated using the formula:

W = PΔV

where:
W is the work done,
P is the pressure, and
ΔV is the change in volume.

The pressure P is given as 0.800 atm, but we need to convert this to pascals (Pa) because the SI unit for work (Joule) is derived from pascals and liters. We know that 1 atm = 101325 Pa, so:

P = 0.800 atm * 101325 Pa/atm = 81060 Pa

The change in volume ΔV is the final volume minus the initial volume:

ΔV = 3.00 L - 6.00 L = -3.00 L

We also need to convert this to cubic meters (m^3) because the SI unit for volume is m^3. We know that 1 L = 0.001 m^3, so:

ΔV = -3.00 L * 0.001 m^3/L = -0.003 m^3

Now we can calculate the work done:

W = PΔV = 81060 Pa * -0.003 m^3 = -243.18 J

So, the work done on the gas is -243.18 J.

(b) The change in internal energy ΔU of the gas can be calculated using the first law of thermodynamics:

ΔU = Q - W

where:
ΔU is the change in internal energy,
Q is the heat transferred to the system, and
W is the work done by the system.

In this case, the heat Q is given as -360 J (the negative sign indicates that energy is leaving the gas), and we've calculated the work W as -243.18 J. So:

ΔU = Q - W = -360 J - (-243.18 J) = -116.82 J

So, the change in internal energy of the gas is -116.82 J.

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