A gas occupies 500mL at 27°C and 1atm — find volume at STP

Question

A gas occupies 500 mL at 27°C and 1 atm pressure. Calculate its volume at STP.

Solution — Step by Step

Given:

$V_1 = 500$ mL = $0.5$ L
$T_1 = 27°C = 27 + 273 = 300$ K
$P_1 = 1$ atm

At STP (Standard Temperature and Pressure):

$T_2 = 0°C = 273$ K
$P_2 = 1$ atm

Find: $V_2$

The Combined Gas Law relates the initial and final states of a fixed amount of ideal gas:

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

Since $P_1 = P_2 = 1$ atm, the pressure terms cancel:

\frac{V_1}{T_1} = \frac{V_2}{T_2}

V_2 = V_1 \times \frac{T_2}{T_1} = 500 \times \frac{273}{300}

V_2 = 500 \times 0.91 = \mathbf{455 \text{ mL}}

Why This Works

STP is colder (273 K) than the initial temperature (300 K). At constant pressure, volume is directly proportional to temperature (Charles’s Law: $V \propto T$ ). Cooling the gas reduces the volume proportionally.

The factor $\frac{T_2}{T_1} = \frac{273}{300} \approx 0.91$ tells us the gas shrinks to about 91% of its original volume when cooled from 300 K to 273 K.

Alternative Method — Using Charles’s Law Directly

Since pressure is constant, this is a pure Charles’s Law problem:

\frac{V_1}{T_1} = \frac{V_2}{T_2}

V_2 = \frac{V_1 T_2}{T_1} = \frac{500 \times 273}{300} = 455 \text{ mL}

Common Mistake

The most common error is using Celsius directly instead of converting to Kelvin. If you write $\frac{500}{27} = \frac{V_2}{0}$ , the equation becomes meaningless (division by zero) because 0°C in Celsius is not “zero temperature.” Always convert: $T(\text{K}) = T(°C) + 273$ . This rule has no exceptions in gas law calculations.

Note: the “old STP” is 0°C and 1 atm (used in older textbooks and most Indian curriculum). The “new IUPAC STP” is 0°C and 100 kPa (≈ 0.987 atm). For CBSE and JEE, always use 0°C and 1 atm unless the question specifies otherwise.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Charles’s Law Directly

Common Mistake

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