Question

A certain quantity of gas has a v of .097L at 298 K At what T would the quantity of gas be reduced to 0.95 l, assuming constant pressure?

Answer 1

The ideal gas law is the equation of state of a hypothetical ideal gas. It is a good approximation to the behavior of many gases under many conditions, although it has several limitations. It was first stated by Émile Clapeyron in 1834 as a combination of Boyle's law and Charles's law. The ideal gas law is often introduced in its common form:

$$PV = nRT$$

where $P$ is the pressure of the gas, $V$ is the volume of the gas, $n$ is the amount of substance of gas (also known as number of moles), $T$ is the temperature of the gas and $R$ is the ideal, or universal, gas constant,

So if $n = \text{const}$ (it is the same quantity of gas), it is possible to write this law for two cases.

$$\mathrm{P}_1\mathrm{V}_1 = nRT_1$$ $$\mathrm{P}_2\mathrm{V}_2 = nRT_2$$

But when $n = \text{const}$ and $R = \text{const}$, and even $P_1 = P_2 = \text{const}$ it is possible to write these two equations together:

$$\mathrm{V}_1 / \mathrm{T}_1 = nR / P_1$$ $$\mathrm{V}_2 / \mathrm{T}_2 = nR / P_2 \text{ if } P_1 = P_2$$ $$\mathrm{V}_1 / \mathrm{T}_1 = \mathrm{V}_2 / \mathrm{T}_2$$ $$\mathrm{V}_1\mathrm{T}_2 = \mathrm{V}_2\mathrm{T}_1$$ $$\mathrm{T}_2 = \mathrm{V}_2\mathrm{T}_1 / \mathrm{V}_1$$

Given:

$$\mathrm{V}_1 = 0.097\mathrm{L}$$ $$\mathrm{V}_2 = 0.95\mathrm{L}$$ $$\mathrm{T}_1 = 298\mathrm{K}$$ $$\mathrm{T}_2 = 0.95 \times 298 / 0.097 = 2919\mathrm{K}$$

New temperature is $2919\mathrm{K}$