Answer:
(c, d)
Work done in isothermal process
can be calculated as:
\[w=2.303nRT\,\log
\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\]
\[\frac{{{w}_{600}}}{{{w}_{300}}}=\frac{2.303\times
1\times R\times 600\log (10)}{2.303\times 1\times R\times 300\log (10)}=2\]
\[\Delta U=0\]for
isothermal processes.
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