A) 352.49 J
B) 849.35 J
C) 249.35 J
D) 302.98 J
Correct Answer: C
Solution :
Key Idea \[W=P\,.\,\,\Delta V\] and \[\Delta E=q+W\] where, \[\Delta E\] is change in internal energy. On absorbing heat, the gas undergoes expansion, and does work on the surrounding, ie, work is done by the system. Thus, the work done (W) is negative. Here, P = 1 arm, \[{{V}_{f}}=2.5\,L\], \[{{V}_{i}}=2\,L,\] \[Q=300\,J\] So, \[W=-P\,.\,\Delta V=-P\,({{V}_{f}}-{{V}_{i}})\] \[=-1\,(2.5-2)\] \[=-0.5\,L-atm\] \[=-0.5\times 101.3J\] \[(\because \,1\,L\,atm=101.3\,J)\] = - 50.65 J Acceding to first law of thermodynamics \[\Delta E=Q+W\] \[\Delta E=300+(-50.65)\] \[\Delta E=249.35\,J\]
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