Two moles of an ideal gas are compressed isot | vedclass

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(A) For an isothermal reversible compression,the work done is given by the formula: $W = -2.303 \ nRT \log_{10} \frac{V_2}{V_1}$ Given: $n = 2 \ mol$,

Vedclass · Accepted Answer

$3.46$

Vedclass · Answer

(A) For an isothermal reversible compression,the work done is given by the formula: $W = -2.303 \ nRT \log_{10} \frac{V_2}{V_1}$ Given: $n = 2 \ mol$,$T = 300 \ K$,$V_1 = 40 \ L$,$V_2 = 20 \ L$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$ Substituting the values: $W = -2.303 	imes 2 	imes 8.314 	imes 300 	imes \log_{10} \frac{20}{40}$ $W = -2.303 	imes 2 	imes 8.314 	imes 300 	imes \log_{10} (0.5)$ Since $\log_{10} (0.5) \approx -0.3010$: $W = -2.303 	imes 2 	imes 8.314 	imes 300 	imes (-0.3010) \approx 3457.97 \ J$ Converting to $kJ$: $W \approx 3.46 \ kJ$ Since work is done on the system,the value is positive.

Two moles of an ideal gas are compressed isothermally and reversibly from $40 \ L$ to $20 \ L$ at $300 \ K$. What is the work done (in $kJ$)? $(R=8.314 \ J \ K^{-1} \ mol^{-1})$

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