If 1.5 L of an ideal gas at a pressure of 20 | vedclass

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(D) For an isothermal reversible expansion,the work done by the gas is given by the formula: $W = -nRT \ln(\frac{V_2}{V_1})$.Since the process is isot

Vedclass · Accepted Answer

$-69.09$

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(D) For an isothermal reversible expansion,the work done by the gas is given by the formula: $W = -nRT \ln(\frac{V_2}{V_1})$.Since the process is isothermal,$p_1 V_1 = nRT$.Given $p_1 = 20 \ atm$ and $V_1 = 1.5 \ L$,we have $nRT = 20 \ atm 	imes 1.5 \ L = 30 \ L \ atm$.Substituting the values into the work formula:$W = -30 	imes \ln(\frac{15}{1.5})$$W = -30 	imes \ln(10)$$W = -30 	imes 2.303 	imes \log_{10}(10)$$W = -30 	imes 2.303 	imes 1 = -69.09 \ L \ atm$.The negative sign indicates that work is done by the system on the surroundings.

If $1.5 \ L$ of an ideal gas at a pressure of $20 \ atm$ expands isothermally and reversibly to a final volume of $15 \ L$,the work done by the gas in $L \ atm$ is

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