A) \[5.75\text{ }J\text{ }{{K}^{-1}}\text{ }mo{{l}^{-1}}\]
B) \[~10.5\text{ }J\text{ }{{K}^{-1}}\text{ }mo{{l}^{-1}}\]
C) \[~21\text{ }J\text{ }{{K}^{-1\text{ }}}mo{{l}^{-1~}}\]
D) \[42\text{ }J\text{ }{{K}^{-1}}mo{{l}^{-1}}\]
Correct Answer: C
Solution :
Since, the volume remains constant, dW= 0. Hence, \[\Delta Q=\Delta U=420\,J.\]But \[\Delta Q=n{{C}_{v}}dT=2\times {{C}_{v}}\times 10\]\[(\because \,dT=10{{\,}^{o}}C=10\,K)\] or \[20{{C}_{v}}=420\,or\] \[{{C}_{v}}=21\,J{{K}^{-1}}mo{{l}^{-1}}\] Hence, the correction option is [c].
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