Question

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a mixture of xenon and argon gas is compressed from a volume of 57.0 l to a volume of 45.0 l, while the pressure is held constant at 13.0 atm. calculate the work done on the gas mixture. round your answer to 3 significant digits, and be sure it has the correct sign (positive or negative).
kj

Explanation:

Step1: Recall the formula for work at constant pressure

The formula for work done on a gas at constant pressure is \( w = -P\Delta V \), where \( \Delta V = V_2 - V_1 \). But when calculating work done on the gas, we can also think in terms of the work done by the gas and then adjust the sign. The work done by the gas is \( w_{by} = -P\Delta V \), so work done on the gas \( w_{on} = P\Delta V \) (wait, no, let's be precise. The first law: \( \Delta U = q + w \), where \( w \) is work done on the system. For a gas expanding against constant pressure, work done by the system is \( w_{by} = P\Delta V \), so work done on the system is \( w = -P\Delta V \)? Wait, no, let's define:

\( \Delta V = V_2 - V_1 \). If the gas is compressed, \( V_2 < V_1 \), so \( \Delta V \) is negative.

Work done by the gas: \( w_{by} = P\Delta V \) (from \( w = \int P dV \), at constant P, \( w = P\Delta V \)). So work done on the gas is \( -w_{by} = -P\Delta V \). Wait, let's check units.

First, calculate \( \Delta V \): \( V_1 = 57.0 \, L \), \( V_2 = 45.0 \, L \), so \( \Delta V = V_2 - V_1 = 45.0 - 57.0 = -12.0 \, L \).

Now, work done by the gas: \( w_{by} = P\Delta V = 13.0 \, atm \times (-12.0 \, L) = -156 \, L \cdot atm \).

Work done on the gas is \( -w_{by} = 156 \, L \cdot atm \).

Now convert \( L \cdot atm \) to joules. The conversion factor is \( 1 \, L \cdot atm = 101.325 \, J \).

So \( 156 \, L \cdot atm \times 101.325 \, J/(L \cdot atm) = 156 \times 101.325 \, J \).

Calculate that: \( 156 \times 101.325 = 156 \times 100 + 156 \times 1.325 = 15600 + 206.7 = 15806.7 \, J \).

Convert to kJ: divide by 1000, so \( 15.8067 \, kJ \). Wait, but let's check the sign again.

Wait, when the gas is compressed, the surroundings do work on the gas, so work done on the gas should be positive. Let's rederive:

The formula for work done on the system (gas) at constant pressure is \( w = -P\Delta V \). Wait, let's use the definition: \( w \) is work done on the system. So if the volume decreases (compression), \( \Delta V = V_2 - V_1 = 45 - 57 = -12 \, L \). Then \( w = -P\Delta V = -13.0 \, atm \times (-12.0 \, L) = 156 \, L \cdot atm \). Yes, that's correct. Because \( \Delta V \) is negative, so \( -P\Delta V \) is positive (since P is positive, and \( \Delta V \) is negative, so negative times negative is positive). So that's work done on the gas.

Now convert \( L \cdot atm \) to J: \( 1 \, L \cdot atm = 101.325 \, J \), so:

\( 156 \, L \cdot atm \times 101.325 \, J/(L \cdot atm) = 156 \times 101.325 \, J \)

Calculate 156 * 101.325:

101.325 * 150 = 15198.75

101.325 * 6 = 607.95

Total: 15198.75 + 607.95 = 15806.7 J

Convert to kJ: 15806.7 J / 1000 = 15.8067 kJ

Round to 3 significant digits: 15.8 kJ? Wait, no, 156 L·atm: 156 has 3 significant digits, 13.0 has 3, 57.0 and 45.0 have 3. So the result should have 3 significant digits. Wait, 156 * 101.325 = 156 * 101.325. Let's do the multiplication more accurately:

101.325 * 156:

101.325 * 100 = 10132.5

101.325 * 50 = 5066.25

101.325 * 6 = 607.95

Sum: 10132.5 + 5066.25 = 15198.75 + 607.95 = 15806.7 J = 15.8067 kJ ≈ 15.8 kJ? Wait, no, wait: \( \Delta V = 45.0 - 57.0 = -12.0 \, L \) (3 sig figs). \( P = 13.0 \, atm \) (3 sig figs). So \( P\Delta V = 13.0 * (-12.0) = -156 \, L·atm \) (3 sig figs). Then work done on the gas is \( - (work done by gas) = - (-156) = 156 \, L·atm \). Then convert to J: 156 * 101.325 = 156 * 101.325. Let's compute 156 * 101.325:

156 * 100 = 15600

156 * 1.325 = 156 * 1 + 156 * 0.3 + 156 * 0.025 = 156 + 46.8 + 3.9 = 206.7

So total is 15600 + 206.7 = 15806.7 J = 15.8067 kJ ≈ 15.8 kJ? Wait, but 101.325 is a conversion factor with more sig figs, so the limiting factor is 156 (3 sig figs), so 15.8 kJ? Wait, no, 156 has 3, 101.325 is exact (conversion factor), so 156 * 101.325 = 15806.7, which is 15.8067 kJ, so 3 sig figs is 15.8 kJ? Wait, but let's check the sign again. Wait, when the gas is compressed, work is done on the gas, so the sign should be positive. Let's confirm the formula:

The work done on the gas (system) at constant pressure is \( w = -P\Delta V \). Wait, no, the standard formula is that work done by the system is \( w_{by} = P\Delta V \), so work done on the system is \( w_{on} = -w_{by} = -P\Delta V \). So \( \Delta V = V_2 - V_1 = 45.0 - 57.0 = -12.0 \, L \). So \( w_{on} = -P\Delta V = -13.0 \, atm * (-12.0 \, L) = 156 \, L·atm \). Then convert to J: 156 L·atm * 101.325 J/(L·atm) = 156 * 101.325 J = 15806.7 J = 15.8067 kJ ≈ 15.8 kJ. Wait, but maybe I made a mistake in the sign. Let's think again: when the gas is compressed, the surroundings push on the gas, so work is done on the gas, so the work should be positive. The volume change is negative (decrease), so \( \Delta V = -12.0 \, L \). Work done by the gas is \( P\Delta V = 13.0 * (-12.0) = -156 \, L·atm \) (negative because the gas is compressed, so work is done on the gas, not by it). So work done on the gas is the negative of work done by the gas: \( - (-156) = 156 \, L·atm \), which is positive, as expected. Then converting to kJ: 156 * 101.325 / 1000 = (156 * 101.325) / 1000. Let's compute that:

156 * 101.325 = 156 * (100 + 1.325) = 15600 + 156*1.325 = 15600 + 206.7 = 15806.7

15806.7 / 1000 = 15.8067 kJ ≈ 15.8 kJ (3 sig figs). Wait, but let's check with another approach.

Alternatively, the formula for work done on the gas when compressed at constant pressure is \( w = P(V_1 - V_2) \), because \( \Delta V = V_2 - V_1 \), so \( V_1 - V_2 = -\Delta V \), so \( w = P(V_1 - V_2) \). Let's use that: \( V_1 - V_2 = 57.0 - 45.0 = 12.0 \, L \). Then \( w = 13.0 \, atm * 12.0 \, L = 156 \, L·atm \), same as before. Then convert to J: 156 * 101.325 = 15806.7 J = 15.8067 kJ ≈ 15.8 kJ. Wait, but 101.325 J/L·atm is the conversion factor, so that's correct.

Wait, but maybe I messed up the conversion. Let's check: 1 L·atm = 101.325 J, so 156 L·atm * 101.325 J/L·atm = 156 * 101.325 J. Let's compute 156 * 101.325:

101.325 * 156:

= (100 + 1.325) * 156

= 100*156 + 1.325*156

= 15600 + 206.7

= 15806.7 J

= 15.8067 kJ ≈ 15.8 kJ (3 significant figures).

So the work done on the gas is +15.8 kJ (or 15.8 kJ, since positive indicates work done on the system).

Wait, but let's check the problem statement: "Calculate the work done on the gas mixture". So the sign should be positive because work is done on the gas. The volume decreases, so the surroundings do work on the gas, so positive work.

So the final answer is 15.8 kJ? Wait, but let's check with more precise calculation. 156 * 101.325 = 156 * 101.325. Let's do 156 * 101.325:

101.325 * 150 = 15198.75

101.325 * 6 = 607.95

Sum: 15198.75 + 607.95 = 15806.7 J = 15.8067 kJ ≈ 15.8 kJ (3 sig figs).

Yes, that seems correct.

Step2: Convert L·atm to joules and then to kilojoules

First, we have \( w = 156 \, L \cdot atm \). The conversion factor from \( L \cdot atm \) to joules is \( 1 \, L \cdot atm = 101.325 \, J \). So:

\( w = 156 \, L \cdot atm \times 101.325 \, \frac{J}{L \cdot atm} = 156 \times 101.325 \, J \)

Calculate \( 156 \times 101.325 \):

\( 156 \times 100 = 15600 \)

\( 156 \times 1.325 = 206.7 \)

\( 15600 + 206.7 = 15806.7 \, J \)

Convert joules to kilojoules by dividing by 1000:

\( w = \frac{15806.7 \, J}{1000} = 15.8067 \, kJ \)

Step3: Round to 3 significant digits

The value \( 15.8067 \, kJ \) rounded to 3 significant digits is \( 15.8 \, kJ \).

Answer:

15.8