Answer:
[tex]117.38^{\circ}C[/tex]
Explanation:
We are given that
Initial temperature=[tex]T_1=127^{\circ}[/tex]
Number of moles=n=5
Heat absorbed,Q=1500 J
Work done=W=2100 J
We have to find the final temperature of the gas.
[tex]\Delta U=Q-W=1500-2100=-600 J[/tex]
[tex]T_2=\frac{2\Delta U}{3nR}+T_1[/tex]
Where R=8.314 J/mol-K
Using the formula
[tex]T_2=\frac{2\times (-600)}{3\times 5\times 8.314}+127^{\circ} C[/tex]
[tex]T_2=117.38^{\circ}C[/tex]
