A certain volume of gas was confined in a rigid container. If the pressure of the gas sample in the container was doubled what happened to the temperature?

Since we know gas law PV =nRT
But there is volume and gas constant is same,
Hence by omission of both from the formula
P = T
Hence the temperature will be in same prportion as pressure.
Hence doubled the pressure, doubled the temperature..

This is an example of Gay-Lussac's law, which states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. In other words, if the pressure goes up, so does the temperature, and vice versa.

The formula for Gay-Lussac's law is:

#P_1/T_1=P_2/T_2#,

where #P_1# is the initial pressure, #P_2# is the final temperature, #T_1# is the initial temperature, and #T_2# is the final temperature.

Example:

A gas inside a rigid container is at #"STP":# #"273 K"# and #"100 kPa"#. If the pressure is doubled, what is the new temperature?

Rearrange the formula to isolate #T_2#. Plug in the known values and solve.

#T_2=(P_2T_1)/P_1#

#T_2=(200.0color(red)cancel(color(black)("kPa"))xx273.15"K")/(100.0color(red)cancel(color(black)("kPa")))="546.3 K"#

#(546.3color(red)cancel(color(black)("K")))/(273.15color(red)cancel(color(black)("K")))=2#