Three moles of an ideal gas expand reversibly | vedclass

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(D) Given: Number of moles of ideal gas,$n = 3$ Initial volume,$V_1 = 2 \, L$ Final volume,$V_2 = 20 \, L$ Temperature,$T = 300 \, K$ For an isotherma

Vedclass · Accepted Answer

$17.2$

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(D) Given: Number of moles of ideal gas,$n = 3$ Initial volume,$V_1 = 2 \, L$ Final volume,$V_2 = 20 \, L$ Temperature,$T = 300 \, K$ For an isothermal reversible process,the change in internal energy $\Delta U = 0$. According to the $1^{st}$ law of thermodynamics,$q = -w$. The heat absorbed $(q)$ is given by: $q = 2.303 n R T \log \frac{V_2}{V_1}$ Substituting the values: $q = 2.303 	imes 3 	imes 8.314 	imes 300 	imes \log \frac{20}{2}$ $q = 2.303 	imes 3 	imes 8.314 	imes 300 	imes \log(10)$ Since $\log(10) = 1$: $q = 2.303 	imes 3 	imes 8.314 	imes 300 	imes 1 \approx 17200 \, J$ Since the question asks for heat change per mole: $q_{mol} = \frac{17200 \, J}{3 \, mol} \approx 5733 \, J/mol = 5.73 \, kJ/mol$. Wait,re-evaluating the calculation: $q = 2.303 	imes 3 	imes 8.314 	imes 300 	imes 1 = 17200 \, J$. If the total heat is $17.2 \, kJ$,then for $3$ moles,the heat per mole is $5.73 \, kJ/mol$. However,if the question implies the total heat for the $3$ moles is the answer,then $17.2$ is correct. Given the options,$17.2$ is the intended answer.

Three moles of an ideal gas expand reversibly under isothermal conditions from $2 \, L$ to $20 \, L$ at $300 \, K$. The amount of heat change (in $kJ / mol$) in the process is:

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