# A gas occupies a volume of 444 mL at 273 K and 79.0 kPa. What is the final kelvin temperature when the volume of the gas is change 880 mL and the pressure is changed to 38.7 kPa?

May 21, 2018

265 K

#### Explanation:

Using the combined gas law you can solve for your missing variable.

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

P1 = 79.0 kPa
V1 = 444 mL
T1 = 273 K

P2 = 38.7 kPa
V2 = 880 mL
T2 = ?

Using the combined gas law we need to rearrange our equation to solve for T2. The first step in this process will be to times both T1 and T2 to both sides of the equation.

$\left(T 1 T 2\right) \frac{P 1 V 1}{T 1} = \left(T 1 T 2\right) \frac{P 2 V 2}{T 2}$

Because T1 and T2 are on both sides of the denominator they cancel each other out, leaving us with the new equation:

$T 2 P 1 V 1 = T 1 P 2 V 2$

We now need to divide P1 and V1 to both sides of the equation and cancel out once again, giving us the final equation:

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

We can now plug in the numbers and solve for T2:

$T 2 = \frac{273 K \cdot 38.7 k P a \cdot 880 m L}{79.0 k P a \cdot 444 m L} = 265 K$

Because kPa and mL are located on both sides of the denominator the units cancel out leaving use the the units K.