Missing question:
"What is the final temperature of the gas?"
Solution:
The first law of thermodynamics states that:
[tex]\Delta U = Q-W[/tex]
where
[tex]\Delta U[/tex] is the variation of internal energy of the gas
[tex]Q[/tex] is the heat absorbed by the gas
[tex]W [/tex] is the work done by the gas
The gas in this problems absorbs [tex]Q=+1300 J[/tex] of heat and it does [tex]W=+2080 J[/tex] of work, so its variation of internal energy is:
[tex]\Delta U = 1300 J - 2080 J = -780 J[/tex]
This means the gas has lost internal energy.
But the variation of internal energy is related to the variation of temperature by:
[tex]\Delta U = \frac{k}{2}nR \Delta T [/tex]
where
k is the number of degrees of freedom (k=3 for a monoatomic gas)
n is the number of moles
R is the gas constant
[tex]\Delta T [/tex] is the variation of temperature of the gas.
Re-arranging the equation and using the variation of internal energy that we found at the previous step, we find:
[tex]\Delta T = \frac{2 \Delta U}{knR}= \frac{2 (-780 J)}{3 (5.0 mol)(8.31 J/mol K)} =-12.5 K [/tex]
Variation of temperatures in Kelvin are equal to variation of temperatures in Celsius, so [tex]\Delta T = -12.5 ^{\circ} C[/tex] and we can now find the final temperature of the gas:
[tex]T_f = T_i + \Delta T=126^{\circ}C - 12.5 ^{\circ}C=113.5^{\circ}C = 386.5 K[/tex]
