# Is gas volume of 444 mL at 273K and 79.0 kPa. What is the final kelvin temperature when the volume of the gas is changed to 1880 mL and the pressure changed to 38.7 kPa?

May 31, 2017

$566 \text{K}$

#### Explanation:

We're asked to calculate the final absolute (Kelvin) temperature of a gas system when it is subdued to known pressure and volume changes.

To solve this problem, we can use the combined gas law:

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

For this equation to work, the units have to be consistent; for example, we can't plug in two values for pressure if one is measured in $\text{kPa}$ and the other in $\text{atm}$; the units must be the same.

Since all the units are consistent, here, solving the problem is straightforward enough: we simply plug in the known variables and solve the equation for the final temperature, ${T}_{2}$:

T_2 = (T_1P_2V_2)/(P_1V_1) = ((273"K")(38.7cancel("kPa"))(1880cancel("mL")))/((79.0cancel("kPa"))(444cancel("mL")))

= color(red)(566"K"

rounded to $3$ significant figures, the number given in the problem.

Thus, the final Kelvin temperature after the changes in pressure and volume is $566 \text{K}$

May 31, 2017

The final temperature will be $\text{566 K}$.

#### Explanation:

This a question involving the Combined Gas Law. You can know that this involves the Combined Gas Law, because the variables are pressure, volume, and temperature, and they are the variables in the Combined Gas Law. The equation used to solve questions involving the Combined Gas Law is:

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

Organize you data.

Known

${P}_{1} = \text{79.0 kPa}$

${V}_{1} = \text{444 mL}$

${T}_{1} = \text{273 K}$

${P}_{2} = \text{38.7 kPa}$

${V}_{2} = \text{1880 mL}$

Unknown: ${T}_{2}$

Solution
Rearrange the equation above to isolate ${T}_{2}$. Insert the known data into the resulting equation and solve.

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

T_2=(38.7color(red)cancel(color(black)("kPa"))xx1880color(red)cancel(color(black)("mL"))xx273"K")/(79.0color(red)cancel(color(black)("kPa"))xx444color(red)cancel(color(black)("mL")))="566 K"