some about themodynamics

Assume that helium gas behaves as an ideal gas, then the following ideal gas equation may be applied: $PV=nRT \quad Eqn (1) \quad$ where P , V, n , R , T refers to pressure, volume, number of moles of the helium gas,gas constant and absolute temperature of the helium gas respectively.

Lets label the containers 1 and 2.

Case a: Container 1 and 2 are at 273K

Case b: Container 1 is at 373K and container 2 is at 273K

Since both containers are in a closed system, the number of moles of gas in both cases must be the same. Note that both containers are at allowed to flow until the system is in equilibrium and at that point pressure is equal in both containers.Rearranging eqn 1 we get:

$n=\frac { PV }{ RT }$

and by using the information that the number of moles of gas must be the same in both cases, we get the following eqn:

$\frac { { P }_{ a }{ V }_{ 1 } }{ { RT }_{ a1 } } +\frac { { P }_{ a }{ V }_{ 2 } }{ R{ T }_{ a2 } } =\frac { { P }_{ b }{ V }_{ 1 } }{ R{ T }_{ b1 } } +\frac { { P }_{ b }{ V }_{ 2 } }{ R{ T }_{ b2 } }$

The R cancels and we plug in the numbers P=1 atm, V=100 cubic cm, and the absolute temperatures for case a and b to get:

$\frac { (1)(100) }{ 273 } +\frac { (1)(100) }{ 273 } =\frac { { P }_{ b }(100) }{ 373 } +\frac { { P }_{ b }(100) }{ 273 }$

and we find that ${P}_{b}=1.15 atm$

Volume remaining same hot gas will have more pressure forcing some gas to flow out to cold container, increasing the pressure of the cold container.
Since the volumes are same, mass in each container =k*P/T. .Where k is appropriate constant in the given situation (which gets canceled out).

Initial total mass = final total mass...> 2 * k * 1/273 = k * P/273 +k * P/373.
Solving for P, P = 1.155