A container with a volume of #12 L# contains a gas with a temperature of #210 K#. If the temperature of the gas changes to #420 K# without any change in pressure, what must the container's new volume be?

2 Answers
Mar 17, 2018

Just apply Charle's law for constant pressure and mas of an ideal gas,

So,we have,

#V/T=k# where, #k# is a constant

So,we putting the initial values of #V# and #T# we get,

#k=12/210#

Now,if new volume is #V'# due to temperature #420K#

Then,we get,

#(V')/420=k=12/210#

So,#V'=(12/210)×420=24L#

Mar 17, 2018

Answer:

The new volume is #24# liters or #24 \ "L"#.

Explanation:

Since there is no change in temperature and the number of moles, we use Charles's law, which states that

#VpropT#

or

#V_1/T_1=V_2/T_2#

Solving for #V_2#, we get:

#V_2=T_2*V_1/T_1#

Plugging in the given values, we find that

#V_2=420color(red)cancelcolor(black)"K"*(12 \ "L")/(210color(red)cancelcolor(black)"K")#

#=24 \ "L"#