# The compression and expansion of gases form the basis of how air is cooled by air conditioners. Suppose 1.55 L of an ideal gas under 6.38 atm of pressure at 20.5°C is expanded to 6.95 L at 1.00 atm. What is the new temperature?

##### 1 Answer

#### Explanation:

The pressure, temperature, and volume of the gas will change when going from its initial state to its final state, which tells you that you must use the combined gas law to find the new *temperature* of the gas.

As you know, when the *number of moles* of gas is **kept constant**, pressure, volume, and temperature have the following relationship

#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)" "# , where

From this point on, the important thing to remember is that the temperature of the gas **must** be expressed in *Kelvin*, so don't forget to convert it from the given degrees Celsius.

So, rearrange that equation to solve for

#T_2 = P_2/P_1 * V_2/V_1 * T_1#

Plug in your values to get

#T_2 = (1.00 color(red)(cancel(color(black)("atm"))))/(6.38color(red)(cancel(color(black)("atm")))) * (6.95 color(red)(cancel(color(black)("L"))))/(1.55color(red)(cancel(color(black)("L")))) * (273.15 + 20.5)"K"#

#T_2 = "206.38 K"#

Rounded to three sig figs, the answer will be

#T_2 = color(green)("206 K")#

If you want, you can express the answer in degrees Celsius

#T_2[""^@"C"] = 206 - 273.15 = -67.2^@"C"#