An ideal gas initially at 30 K undergoes an isobaric expansion at 2.50 k Pa. If the volume increases from `1.00 m^(2)` to `3.00 m^(3)` and 12.5 kJ is transferred to the gas by heat, what are (a) the change in its internal energy and (b) its final temperature ?

Explanation

The gas pressure is much less than one atmosphere. This could be managed by having the gas in a cylinder with a piston on the bottom end, supporting a constant load that hangs from the piston. The (large) cylineder is put into a warmer environment of make the gas expand.
The first law of thermodynamics will tell us the change in internal energy. The ideal gas law will tells us the change in internal energy. The ideal gas law will tells us the final temperature.
a. `Delta U - Q + W`
where `W = - P Delta V` for a constant pressure process.
So `Delta U = Q - P Delta V`
`Delta U = 1.25 xx 10^(4) J - (2.50 xx 10^(3) N//m^(2))`
`xx (3.00 m^(3) xx 10 m^(3)) = 7500 J`
b. Since pressure and quantity of the gas are constant, we have from the equation of state
`(V_(1))/(T_(1)) = (V_(2))/(T_(2))`
and `T_(2) = ((V_(2))/(V_(1))) T_(1) = ((3.00 m^(3))/(1.00 m^(3))) (300 K) = 900 K`