A) 189 K & + 2700.23 J
B) 100 K & + 2500J
C) 100 K & - 2000,23J
D) 189 K & - 2767.23 J
Correct Answer: D
Solution :
(i) In adiabatic process, \[{{T}_{1}}V_{1}^{\gamma -1}\,={{T}_{2}}V_{2}^{\gamma -1}\,\] and here\[\gamma =5/3\] \[\therefore \,\,300\,\times \,{{V}^{2/3}}\,=T{{(2V)}^{2/3}}\] \[T=300/{{(2)}^{2/3}}\,=189K\] (ii) \[dU=\frac{nRdT}{(\gamma -1)}\,[dU\,=n{{C}_{v}}dT\] and \[{{C}_{v}}=\frac{R}{\gamma -1}\,\]] \[\therefore \,\,dU=2\,\times 3/2\,\times 8.31\,(189-300\,)=2767.23J\]
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