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A gas syringe carried into the cold

SCH3U Grade 11 Chemistry · Gases and Atmospheric Chemistry

At the bench a gas syringe is flushed and filled with dry carbon dioxide until the plunger stands at 500.0 mL, the wall instruments reading 101.3 kPa and 25.0 °C. The nozzle is capped, the seal is checked with a gentle push on the plunger, and a student carries the syringe down the service corridor and out onto a January loading dock. In the cold the plunger is left to find its own level, sliding freely along the barrel until the gas inside has settled to the outdoor conditions: the dock barometer reads 85.0 kPa and its thermometer −15.0 °C. Nothing has entered or left the sealed barrel on the trip; only the state of the gas has changed. Taking CO₂ as 44.01 g/mol, find the volume outdoors, the amount of gas in the syringe, the volume that amount would occupy at STP, and its mass.

500.0 mL · 101.3 kPa · 25.0 °Csealed 0.899 g CO₂515.9 mL · 85.0 kPa · −15.0 °C

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • V₁ = 500 mL — Syringe volume in the lab
  • P₁ = 101.3 kPa — Lab pressure
  • T₁ = 25 °C — Lab temperature
  • P₂ = 85 kPa — Barometer on the loading dock
  • T₂ = -15 °C — Temperature on the loading dock
  • M = 44.01 g/mol — Molar mass of CO₂
Determine
  1. (a)the volume of the gas outdoors
  2. (b)the amount of gas in the syringe
  3. (c)the volume that amount would occupy at STP
  4. (d)the mass of the gas
Step 1 of 4(a) · solve for Final volume

Nothing enters or leaves the syringe, so PV/T is the same at both ends. Lower pressure swells the gas, colder air shrinks it, and the pressure wins — about 516 mL.

P1V1T1P2T2V2
Rearranged for V₂
V2=P1V1 T2T1 P2V_2 = \frac{P_1 V_1 \, T_2}{T_1 \, P_2}
Your values, in your units
V2=(101.3 kPa) (500 mL) (−15 ∘C)(25 ∘C) (85 kPa)V_2 = \frac{\left( 101.3\ \text{kPa} \right) \, \left( 500\ \text{mL} \right) \, \left( -15\ ^{\circ}\text{C} \right)}{\left( 25\ ^{\circ}\text{C} \right) \, \left( 85\ \text{kPa} \right)}
Converted to base units
V2=(101.3 kPa) (0.5 L) (−15 ∘C)(25 ∘C) (85 kPa)V_2 = \frac{\left( 101.3\ \text{kPa} \right) \, \left( 0.5\ \text{L} \right) \, \left( -15\ ^{\circ}\text{C} \right)}{\left( 25\ ^{\circ}\text{C} \right) \, \left( 85\ \text{kPa} \right)}
Answer
V2=515.94 mLV_2 = 515.94\ \text{mL}

Carried onward at full precision, not this rounded figure.

Open the Combined Gas Law solver →

Step 2 of 4(b) · solve for Amount

Now count the gas, using the outdoor state and the volume from step 1. The combined gas law could never have told you this — it only ever compares two states of the same unknown amount.

PVTn
Rearranged for n
n=PVRTn = \tfrac{P V}{R T}
515.94 mLcarried from step 1
Your values, in your units
n=(85 kPa) (0.000515938 m3)(8.31446 J/(mol⋅K)) (−15 ∘C)n = \tfrac{\left( 85\ \text{kPa} \right) \, \left( 0.000515938\ \text{m}^{3} \right)}{\left( 8.31446\ \text{J/(mol}{\cdot}\text{K)} \right) \, \left( -15\ ^{\circ}\text{C} \right)}
Converted to base units
n=(85 kPa) (0.515938 L)(8.31446 J/(mol⋅K)) (−15 ∘C)n = \tfrac{\left( 85\ \text{kPa} \right) \, \left( 0.515938\ \text{L} \right)}{\left( 8.31446\ \text{J/(mol}{\cdot}\text{K)} \right) \, \left( -15\ ^{\circ}\text{C} \right)}
Answer
n=20.432 mmoln = 20.432\ \text{mmol}

Carried onward at full precision, not this rounded figure.

Open the Ideal Gas Law solver →

Step 3 of 4(c) · solve for Gas volume at STP

The same amount reported at STP, about 458 mL — the conventional way to quote a gas sample so that two labs on different days can compare. Nothing downstream needs it.

Vn
Rearranged for V
V=n VmV = n \, V_m
20.432 mmolcarried from step 2
Your values, in your units
V=(0.020432 mol) VmV = \left( 0.020432\ \text{mol} \right) \, V_m
Answer
V=457.96 mLV = 457.96\ \text{mL}

Carried onward at full precision, not this rounded figure.

Open the Gas Volume at STP solver →

Step 4 of 4(d) · solve for Mass

Moles times molar mass. The syringe weighs the same on the loading dock as it did in the lab, which is the sanity check this whole chain is built around.

mMn
Rearranged for m
m=n Mm = n \, M
20.432 mmolcarried from step 2
Your values, in your units
m=(0.020432 mol) (44.01 g/mol)m = \left( 0.020432\ \text{mol} \right) \, \left( 44.01\ \text{g/mol} \right)
Answer
m=899.21 mgm = 899.21\ \text{mg}

Carried onward at full precision, not this rounded figure.

Open the Moles from Mass (n = m/M) solver →

Answer

Therefore the plunger settles at 515.9 mL on the dock, the syringe holds 0.02043 mol of CO₂ — which would stand at 458.0 mL if quoted at STP — and the same 0.899 g of gas rides inside at both ends of the trip.

Why this order

The order here encodes a distinction students routinely miss. The combined gas law is a comparison: it relates two states of one sealed sample and can never produce an amount, because the moles cancel out of it entirely. The ideal gas law is an inventory: it names how much gas there is, but only if you feed it one complete state. So step 1 has to run before step 2, and step 2 has to be fed a matched trio — the outdoor pressure with the outdoor volume and the outdoor temperature. Mixing the indoor pressure with the outdoor volume is the single commonest wreck in this topic.

There is a hidden consistency worth checking by hand. Because PV/T is constant, P₂V₂/T₂ equals P₁V₁/T₁, so n could have been computed from the indoor state without step 1 at all — and it gives the identical 0.02043 mol, which is exactly what conservation of matter demands. If your two routes disagree, the arithmetic is wrong, not the chemistry. The other trap is temperature: −15.0 °C is 258.15 K, and every one of these equations divides by absolute temperature. Feeding a gas law a negative Celsius number produces a negative volume, which at least announces itself; feeding it 25 instead of 298.15 produces a merely wrong answer, which does not.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.