# A container with a volume of #6 L# contains a gas with a temperature of #270^o K#. If the temperature of the gas changes to #330 ^o K# without any change in pressure, what must the container's new volume be?

##### 1 Answer

The new volume is

#### Explanation:

Given that the pressure of the gas remains constant, we can identify this as an *isobaric* process. This is a process is one in which pressure remains constant, but temperature and volume change. We can use Charles' Law to find an unknown temperature or volume value for an isobaric process, given that we know something about the other three values.

#(V_i)/(T_i)=(V_f)/(T_f)#

This law describes how gas expands as temperature increases. The inverse is also true. You don't need to memorize Charles' Law (or Boyle's' or Gay-Lussac's) if you remember the combined gas law, which relates all of these simple gas laws:

#(PV)/T=# constantor, equivalently:

#(P_iV_i)/(T_i)=(P_fV_f)/(T_f)#

( also written

We can see that if pressure is constant,

We want to solve for

#V_f=(V_iT_f)/T_i#

Given

#V_f=(6L*330cancelK)/(270cancelK)#

#V_f=7 1/3L~~7L#