Work of 500 J is done by an ideal gas, when 500 cal of heat is added to the gas. What is the increase in the internal energy of the gas in J?
500 cal=2092 J500\ cal=2092\ J500 cal=2092 J
Q=ΔU+W→ΔU=Q−W=2092−500=1592 (J)Q=\Delta U+W\to \Delta U=Q-W=2092-500=1592\ (J)Q=ΔU+W→ΔU=Q−W=2092−500=1592 (J) . Answer
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments