A rigid container of O2 has a pressure of 340 kPa at a temperature of 713 K. What is the pressure at 273 K if the volume and number of moles has not changed?

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Answer 1 · 2018-02-26T16:23:48

$130.2 kPa$ $130.2"kPa"$

According to Gay-Lussac's Law:

$P \propto T$ $PpropT$ .

It follows that $P = k T$ $P=kT$ , and so:

$\frac{P}{T} = k$ $P/T=k$ . It means that we can also say:

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$ $P_1/T_1=P_2/T_2$

We need to find $P_{2}$ $P_2$ . Rearranging to solve for such:

$P_{2} = \frac{P_{1} T_{2}}{T_{1}}$ $P_2=(P_1T_2)/T_1$

We know that:

$P_{1} = 340 kPa$ $P_1=340"kPa"$

$T_{1} = 713 K$ $T_1=713"K"$

$T_{2} = 273 K$ $T_2=273"K"$

Inputting:

$P_{2} = \frac{340 \cdot 273}{713}$ $P_2=(340*273)/713$

$P_{2} = 130.2 kPa$ $P_2=130.2"kPa"$ is the new pressure.