A) 180 cal
B) 300 cal
C) 360 cal
D) 540 cal
Correct Answer: D
Solution :
Consider \[n\] moles of a gas which undergo isochoric process, i.e., V = constant. From first law of thermodynamics, \[\Delta Q=\Delta W+\Delta U\] ... (i) Here, \[\Delta W=0\] as V = constant \[\Delta Q=n{{C}_{V}}\Delta T\] Substituting in Eq. (i), we get \[\Delta U=n{{C}_{V}}\Delta T\] ... (ii) Mayors formula can be written as \[{{C}_{P}}-{{C}_{V}}=R\] \[\Rightarrow \] \[{{C}_{V}}={{C}_{P}}-R\] ... (ii) From Eqs. (ii) and (iii), we have \[\Delta U=n({{C}_{P}}-R)\Delta T\] Given, \[n=6,\,{{C}_{P}}=8\,cal/mol-K\], \[R=8.31\,J/mol-K\] \[\approx 2\,cal/mol-K\] Hence, \[\Delta U=6(8-2)\,(35-20)\] \[=6\times 6\times 15\] = 540 calYou need to login to perform this action.
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