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About This Drill
ACT Science: Data Representation (Drill 3) is a Science practice drill covering Data Representation. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
ACT Data Representation questions test your ability to extract and apply data from scientific figures. This drill uses a pressure-volume curve and a pressure-temperature table for a gas sample, asking you to identify trends, apply gas law relationships, and evaluate a student's claim.
Questions & Explanations
Figure 1 and Table 1
A chemistry student investigated the relationships between the pressure, volume, and temperature of a fixed amount of an ideal gas (0.10 mol of nitrogen gas, N₂). In Trial 1, the temperature was held constant at 300 K while the volume of the gas was varied from 1.0 L to 5.0 L, and the resulting pressure was recorded (Figure 1). In Trial 2, the volume was held constant at 2.0 L while the temperature was varied from 200 K to 600 K, and the resulting pressure was recorded (Table 1). All measurements were taken at conditions far from N₂'s liquefaction point.
Figure 1. Gas Pressure vs. Volume (T = 300 K)
Pressure (atm)
Volume (L)
0
1
2
3
4
5
1.0
2.0
3.0
4.0
5.0
(1.0 L, 2.46 atm)
(2.0 L, 1.23 atm)
(3.0 L, 0.82 atm)
Table 1. Gas Pressure vs. Temperature (V = 2.0 L)
| Temperature (K) | Pressure (atm) |
|---|
| 200 | 0.82 |
| 300 | 1.23 |
| 400 | 1.64 |
| 500 | 2.05 |
| 600 | 2.46 |
Question 1. According to Figure 1, as the volume of the gas increased from 1.0 L to 5.0 L at constant temperature, the pressure:
-
A) increased linearly
-
B) decreased linearly
-
C) decreased, with the greatest change occurring at smaller volumes ✓
-
D) remained constant
Explanation: Figure 1 shows a curved (hyperbolic) decrease; this is Boyle's Law (P ∝ 1/V at constant T). The curve is steep at low volumes and flattens at higher volumes, meaning the largest pressure drops occur when volume is smallest. Going from 1.0 L to 2.0 L drops pressure by about 1.23 atm; going from 4.0 L to 5.0 L drops it by far less. This rules out a linear decrease and confirms C.
Table 1
A chemistry student investigated the relationships between the pressure, volume, and temperature of a fixed amount of an ideal gas (0.10 mol of nitrogen gas, N₂). In Trial 1, the temperature was held constant at 300 K while the volume of the gas was varied from 1.0 L to 5.0 L, and the resulting pressure was recorded (Figure 1). In Trial 2, the volume was held constant at 2.0 L while the temperature was varied from 200 K to 600 K, and the resulting pressure was recorded (Table 1). All measurements were taken at conditions far from N₂'s liquefaction point.
Table 1. Gas Pressure vs. Temperature (V = 2.0 L)
| Temperature (K) | Pressure (atm) |
|---|
| 200 | 0.82 |
| 300 | 1.23 |
| 400 | 1.64 |
| 500 | 2.05 |
| 600 | 2.46 |
Question 2. According to Table 1, what was the increase in pressure for each 100 K increase in temperature?
-
A) 0.41 atm ✓
-
B) 0.82 atm
-
C) 1.23 atm
-
D) 1.64 atm
Explanation: From Table 1, each 100 K increase produces a constant pressure rise: 200→300 K gives 1.23−0.82 = 0.41 atm; 300→400 K gives 1.64−1.23 = 0.41 atm; and so on. The answer is 0.41 atm per 100 K, a perfectly linear relationship known as Gay-Lussac's Law (P ∝ T at constant V). Gas molecules move faster at higher temperatures and strike container walls more forcefully and frequently, increasing pressure proportionally.
Figure 1 and Table 1
A chemistry student investigated the relationships between the pressure, volume, and temperature of a fixed amount of an ideal gas (0.10 mol of nitrogen gas, N₂). In Trial 1, the temperature was held constant at 300 K while the volume of the gas was varied from 1.0 L to 5.0 L, and the resulting pressure was recorded (Figure 1). In Trial 2, the volume was held constant at 2.0 L while the temperature was varied from 200 K to 600 K, and the resulting pressure was recorded (Table 1). All measurements were taken at conditions far from N₂'s liquefaction point.
Figure 1. Gas Pressure vs. Volume (T = 300 K)
Pressure (atm)
Volume (L)
0
1
2
3
4
5
1.0
2.0
3.0
4.0
5.0
Table 1. Gas Pressure vs. Temperature (V = 2.0 L)
| Temperature (K) | Pressure (atm) |
|---|
| 200 | 0.82 |
| 300 | 1.23 |
| 400 | 1.64 |
| 500 | 2.05 |
| 600 | 2.46 |
Question 3. Based on Figure 1 and Table 1, at a temperature of 300 K and a volume of 2.0 L, what was the pressure of the gas?
-
A) 0.82 atm
-
B) 1.23 atm ✓
-
C) 1.64 atm
-
D) 2.46 atm
Explanation: Both Figure 1 and Table 1 confirm this data point. Figure 1 is plotted at T = 300 K, and the labeled point at 2.0 L reads 1.23 atm. Table 1 independently confirms this, at 300 K with volume fixed at 2.0 L, the recorded pressure is 1.23 atm. This is also consistent with the ideal gas law: P = nRT/V = (0.10)(0.0821)(300)/2.0 ≈ 1.23 atm.
Table 1
A chemistry student investigated the relationships between the pressure, volume, and temperature of a fixed amount of an ideal gas (0.10 mol of nitrogen gas, N₂). In Trial 1, the temperature was held constant at 300 K while the volume of the gas was varied from 1.0 L to 5.0 L, and the resulting pressure was recorded (Figure 1). In Trial 2, the volume was held constant at 2.0 L while the temperature was varied from 200 K to 600 K, and the resulting pressure was recorded (Table 1). All measurements were taken at conditions far from N₂'s liquefaction point.
Table 1. Gas Pressure vs. Temperature (V = 2.0 L)
| Temperature (K) | Pressure (atm) |
|---|
| 200 | 0.82 |
| 300 | 1.23 |
| 400 | 1.64 |
| 500 | 2.05 |
| 600 | 2.46 |
Question 4. Based on Table 1, what would be the most likely pressure of the gas at 700 K and V = 2.0 L?
-
A) 2.46 atm
-
B) 2.87 atm ✓
-
C) 3.28 atm
-
D) 4.10 atm
Explanation: Table 1 shows a constant increase of 0.41 atm per 100 K. Extrapolating one step beyond 600 K: 2.46 + 0.41 = 2.87 atm at 700 K. This is valid for an ideal gas because P is directly proportional to T at constant volume (Gay-Lussac's Law), and the uniform increments in the table confirm the relationship remains linear throughout.
Figure 1
A chemistry student investigated the relationships between the pressure, volume, and temperature of a fixed amount of an ideal gas (0.10 mol of nitrogen gas, N₂). In Trial 1, the temperature was held constant at 300 K while the volume of the gas was varied from 1.0 L to 5.0 L, and the resulting pressure was recorded (Figure 1). In Trial 2, the volume was held constant at 2.0 L while the temperature was varied from 200 K to 600 K, and the resulting pressure was recorded (Table 1). All measurements were taken at conditions far from N₂'s liquefaction point.
Figure 1. Gas Pressure vs. Volume (T = 300 K)
Pressure (atm)
Volume (L)
0
1
2
3
4
5
1.0
2.0
3.0
4.0
5.0
Question 5. A student claims that doubling the volume of a gas at constant temperature will halve its pressure. Is this claim consistent with Figure 1?
-
A) Yes; when volume doubles from 1.0 L to 2.0 L, pressure drops from approximately 2.46 atm to 1.23 atm ✓
-
B) Yes; when volume doubles from 2.0 L to 4.0 L, pressure drops from approximately 1.64 atm to 0.82 atm
-
C) No; pressure decreases by the same fixed amount each time volume doubles
-
D) No; pressure increases when volume doubles
Explanation: From Figure 1 at T = 300 K: at 1.0 L pressure is approximately 2.46 atm, and at 2.0 L it is approximately 1.23 atm, exactly half. This confirms Boyle's Law (P₁V₁ = P₂V₂). Option B cites values of 1.64 and 0.82 atm, which come from the Table 1 dataset (T = 200 K, not 300 K) and do not appear on Figure 1. Option A correctly reads Figure 1 at the 1.0→2.0 L doubling.