At 27 , ^oC,one mole of an ideal gas is compr | vedclass

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(B) For an isothermal process involving an ideal gas,the internal energy change $\Delta E = 0$ because $\Delta E$ depends only on temperature.Accordin

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$-965.84 \ cal, +965.84 \ cal$

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(B) For an isothermal process involving an ideal gas,the internal energy change $\Delta E = 0$ because $\Delta E$ depends only on temperature.According to the first law of thermodynamics,$\Delta E = q + w$,so $q = -w$.The work done in a reversible isothermal compression is given by $w = -2.303 \ nRT \ \log \frac{P_2}{P_1}$.Given $n = 1 \ mol$,$R = 2 \ cal \ K^{-1} \ mol^{-1}$,$T = 27 + 273 = 300 \ K$,$P_1 = 2 \ atm$,and $P_2 = 10 \ atm$.$w = -2.303 	imes 1 	imes 2 	imes 300 \ \log \frac{10}{2} = -2.303 	imes 600 	imes \log 5 = -1381.8 	imes 0.699 = -965.84 \ cal$.Since $q = -w$,$q = -(-965.84 \ cal) = +965.84 \ cal$.Therefore,$\Delta E = 0$ and $q = +965.84 \ cal$.

At $27 \, ^oC$,one mole of an ideal gas is compressed isothermally and reversibly from a pressure of $2 \ atm$ to $10 \ atm$. The values of $\Delta E$ and $q$ are $(R = 2 \ cal \ K^{-1} \ mol^{-1})$:

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