Thermodynamics Question 292
Question: Suppose 0.5 mole of an ideal gas undergoes an isothermal expansion as energy is added to its heat Q. Graph shows the final volume $ V _{f} $ versus Q. The temperature of the gas is (use ln 9 = 2 and$ R=\frac{25}{3}J/mol-K $ )
Options:
A) 360 K
B) 293 K
C) 386 K
D) 412 K
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ Q=W=nRT\text{ in }\frac{V _{f}}{V _{i}} $
$ T=\frac{Q}{nR\text{ In}( V _{f}/V _{i} )}=\frac{1500}{0.5\times 25/3\times \text{ ln 3}} $
$ \Rightarrow T=\frac{1500}{0.5\times 25/3\times \text{1}}=360K $
