The work done to double the volume of 0.1 mo | vedclass

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(C) At constant pressure,according to Charles's Law,$\frac{V_1}{T_1} = \frac{V_2}{T_2}$.Given $V_1 = V$,$V_2 = 2V$,and $T_1 = 27 + 273 = 300 \ K$.Subs

Vedclass · Accepted Answer

$60$

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(C) At constant pressure,according to Charles's Law,$\frac{V_1}{T_1} = \frac{V_2}{T_2}$.Given $V_1 = V$,$V_2 = 2V$,and $T_1 = 27 + 273 = 300 \ K$.Substituting the values: $\frac{V}{300} = \frac{2V}{T_2}$,which gives $T_2 = 600 \ K$.The change in temperature is $\Delta T = T_2 - T_1 = 600 - 300 = 300 \ K$.The work done in an isobaric process is given by $W = P \Delta V = \mu R \Delta T$.Using $\mu = 0.1 \ mol$ and $R \approx 2 \ cal/(mol \cdot K)$:$W = (0.1) 	imes (2) 	imes (300) = 60 \ cal$.

The work done to double the volume of $0.1 \ mol$ of a gas at constant pressure and a temperature of $27 \ ^\circ C$ is equal to ..... $cal$.

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