In an isothermal and reversible process,1.6 t | vedclass

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(D) Given: $V_1 = 10 \ dm^3, V_2 = 100 \ dm^3, T = 300 \ K, R = 8.314 \ J \ K^{-1} \ mol^{-1}$.Mass of $O_2$ $(m)$ = $1.6 	imes 10^{-2} \ kg = 16 \ g

Vedclass · Accepted Answer

$-2872$

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(D) Given: $V_1 = 10 \ dm^3, V_2 = 100 \ dm^3, T = 300 \ K, R = 8.314 \ J \ K^{-1} \ mol^{-1}$.Mass of $O_2$ $(m)$ = $1.6 	imes 10^{-2} \ kg = 16 \ g$.Molar mass of $O_2$ $(M)$ = $32 \ g \ mol^{-1}$.Number of moles $(n)$ = $\frac{m}{M} = \frac{16}{32} = 0.5 \ mol$.For an isothermal and reversible expansion process,the work done is given by:$W = -2.303 \ nRT \ \log_{10} \left( \frac{V_2}{V_1} ight)$.Substituting the values:$W = -2.303 	imes 0.5 	imes 8.314 	imes 300 	imes \log_{10} \left( \frac{100}{10} ight)$.$W = -2.303 	imes 0.5 	imes 8.314 	imes 300 	imes \log_{10} (10)$.Since $\log_{10} (10) = 1$,$W = -2.303 	imes 0.5 	imes 8.314 	imes 300 = -2872 \ J$.

In an isothermal and reversible process,$1.6 \times 10^{-2} \ kg$ of $O_2$ expands from $10 \ dm^3$ to $100 \ dm^3$ at $300 \ K$. The work done in the process is: $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$ (in $J$)

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