A gas has a pressure of 710 kPa at 227°C. What will its pressure be at 27.0°C, if the volume doubles?

Jun 15, 2017

$\text{Final Pressure} = 213$ $\text{kPa}$

Explanation:

We're asked to calculate the final pressure of the gas, given its change in temperature and the fact that its volume doubled.

We can solve this problem using the combined gas law:

$\frac{{P}_{1} {V}_{1}}{{T}_{1}} = \frac{{P}_{2} {V}_{2}}{{T}_{2}}$

We must convert our given temperatures to Kelvin, which we always must do for gas equations:

${T}_{1} = {227}^{\text{o""C}} + 273 = 500$ $\text{K}$

${T}_{2} = {27.0}^{\text{o""C}} + 273 = 300$ $\text{K}$

We're not given any values for the volume, but we know that ${V}_{2}$ is twice that of ${V}_{1}$, let's make things as simple as possible by making ${V}_{1}$ $1$ liter and ${V}_{2}$ $2$ liters.

Plugging in our known values, and rearranging the equation to solve for the final pressure, ${P}_{2}$, we have

P_2 = (P_1V_1T_2)/(T_1V_2) = ((710"kPa")(1cancel("L"))(300cancel("K")))/((500cancel("K"))(2cancel("L"))) = color(red)(213 color(red)("kPa"

The final pressure is thus color(red)(213 kilopascals