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(B) Given the condition $V \propto T^{2/3}$,we can write $V = k T^{2/3}$.From the ideal gas law $PV = nRT$,we have $P = \frac{nRT}{V}$.Substituting $V

Vedclass · Accepted Answer

$167$

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(B) Given the condition $V \propto T^{2/3}$,we can write $V = k T^{2/3}$.From the ideal gas law $PV = nRT$,we have $P = \frac{nRT}{V}$.Substituting $V$,we get $P = \frac{nRT}{k T^{2/3}} = \frac{nR}{k} T^{1/3}$.Since $T \propto V^{3/2}$,we have $P \propto (V^{3/2})^{1/3} = V^{1/2}$.This is a polytropic process $PV^x = 	ext{constant}$ where $P V^{-1/2} = 	ext{constant}$,so $x = -1/2$.The work done in a polytropic process is given by $W = \frac{nR \Delta T}{1-x}$.Here $n = 1 \ mol$,$\Delta T = 30 \ K$,$R = 1.99 \ cal/mol-K$,and $x = -1/2$.$W = \frac{1 	imes 1.99 	imes 30}{1 - (-1/2)} = \frac{59.7}{1.5} = 39.8 \ cal$.Converting to Joules $(1 \ cal \approx 4.184 \ J)$: $W = 39.8 	imes 4.184 \approx 166.5 \ J$.Rounding to the nearest integer,we get $167 \ J$.

Find the amount of work done to increase the temperature of one mole of an ideal gas by $30^o\ C$ if it is expanding under the condition $V \propto T^{2/3}$. $[R = 1.99 \ cal/mol-K]$

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