A rigid steel cylinder of fixed volume 8.00 dm³ contains carbon dioxide gas, CO₂ (Mr = 44.0), at a pressure of 3.50×10^5 Pa and a temperature of 291 K. Assume that the CO₂ behaves as an ideal gas throughout. (a) Convert the volume into the SI unit required for the ideal gas equa…

AS Chemistry 9701: Using pV=nRT for a Sealed Gas Cylinder (Moles, Mass and Temperature)

A rigid steel cylinder of fixed volume 8.00 dm³ contains carbon dioxide gas, CO₂ ( $M_r = 44.0$ ), at a pressure of $3.50\times10^{5}\ \text{Pa}$ and a temperature of 291 K. Assume that the CO₂ behaves as an ideal gas throughout.

(a) Convert the volume into the SI unit required for the ideal gas equation, and use $pV = nRT$ , where $R = 8.31\ \text{J K}^{-1}\text{mol}^{-1}$ , to calculate the amount, in mol, of CO₂ present in the cylinder. [3]

(b) Calculate the mass, in g, of CO₂ present in the cylinder. [2]

(c) The sealed cylinder is left in direct sunlight and warms up, so that the pressure rises to $4.20\times10^{5}\ \text{Pa}$ while the volume of the cylinder and the amount of gas inside it remain unchanged. Use the ideal gas equation to calculate the new temperature of the gas, in K. [3]

Part (a): Converting volume and calculating the amount of CO₂

The ideal gas equation requires volume in cubic metres ( $\text{m}^3$ ). Since $1\ \text{m}^3 = 10^3\ \text{dm}^3$ : $V = 8.00\ \text{dm}^3 = \frac{8.00}{10^3}\ \text{m}^3 = 8.00\times10^{-3}\ \text{m}^3$

The pressure and temperature are already in the correct SI units: $p = 3.50\times10^{5}\ \text{Pa}$ , $T = 291\ \text{K}$ .

Rearranging $pV = nRT$ for $n$ : $n = \frac{pV}{RT}$

Substituting the values: $n = \frac{(3.50\times10^{5})\times(8.00\times10^{-3})}{8.31\times291}$

Working out the numerator and denominator separately: $pV = 3.50\times10^{5}\times8.00\times10^{-3} = 2800$ $RT = 8.31\times291 = 2418.21$

So: $n = \frac{2800}{2418.21} = 1.15788\ \text{mol}$

(Check: $2418.21\times1.158 \approx 2800.3$ , confirming $n \approx 1.158\ \text{mol}$ before rounding.)

To 3 significant figures, $n = 1.16\ \text{mol}$ .

Part (b): Calculating the mass of CO₂

Using $n = \dfrac{m}{M_r}$ , rearranged for mass: $m = n\times M_r = 1.15788\times44.0$

$m = 50.9\ \text{g}$

(Check: $1.15788\times44.0 = 1.15788\times40 + 1.15788\times4 = 46.315 + 4.632 = 50.947$ , which rounds to $50.9\ \text{g}$ to 3 s.f.)

Part (c): Calculating the new temperature after heating

The cylinder is rigid (so $V$ is unchanged) and sealed (so $n$ is unchanged). Since $R$ is a constant, rearranging $pV = nRT$ gives: $\frac{p}{T} = \frac{nR}{V} = \text{constant}$

So $\dfrac{p_1}{T_1} = \dfrac{p_2}{T_2}$ , which rearranges to: $T_2 = T_1\times\frac{p_2}{p_1}$

Substituting $T_1 = 291\ \text{K}$ , $p_1 = 3.50\times10^{5}\ \text{Pa}$ and $p_2 = 4.20\times10^{5}\ \text{Pa}$ : $T_2 = 291\times\frac{4.20\times10^{5}}{3.50\times10^{5}} = 291\times1.20$

$T_2 = 349.2\ \text{K}$

To 3 significant figures, $T_2 = 349\ \text{K}$ .

(Check: this is a sensible result. The pressure increased by a factor of 1.20, so the temperature must also increase by exactly the same factor since $V$ and $n$ are fixed, and $291\times1.20 = 349.2$ .)

Final answers

(a) $V = 8.00\times10^{-3}\ \text{m}^3$ , $n = 1.16\ \text{mol}$ (3 s.f.)
(b) $m = 50.9\ \text{g}$ (3 s.f.)
(c) $T_2 = 349\ \text{K}$ (3 s.f.)