spec. heat capacity +ideal gas

Yes, you can.

Using the ideal gas equation PV = nRT
No. of moles n1 is the first container with temperature 310 K is
n1 = (1.2x10^5) x (2x10^-2)/R(310) = 7.742/R
No. of moles n2 in the second container with temp 320 K is
n2 = (1.2x10^5) x (2x10^-2)/R(320) = 7.5/R
Hence, total no. of moles of gas n is
n = n1 + n2 = 7.742/R + 7.5/R = 15.242/R

Using the heat balance equation: heat loss = heat gain
(n2).Cv.(320-T) = (n1).Cv.(T-310)
where Cv is the specific heat capacity at constant volume of the gas, and T is the final temperature.
hence, (7.5/R).Cv.(320-T) = (7.742/R).Cv.(T-310)
i.e. 7.5(320-T) = 7.742(T-310)
T = 315 K

You could also solve the problem using the ideal gas equation.
After the tube is opened, PV = nRT
(1.2x10^5).(2 x 2x10^-2) = (15.242/R).RT
i.e. T =(1.2x10^5 x 4x10^-2)/15.242 K = 315 K

The same result of temperature is obtained.



2012-10-12 15:08:26 補充：
Your suppl question:
The formula Q=mcΔT is generally true for calculating the internal energy of an ideal gas. But you need to be aware that in some thermodynamics process, there is work done by the gas, which should also be taken into accout.