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(A) For an isothermal process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta

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$3.6 	imes 10^3 \ 	ext{cal}$ of heat flowed out from the gas

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(A) For an isothermal process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$. Since $\Delta U = 0$,we have $\Delta Q = \Delta W$. In this case,work is done on the gas,so $\Delta W = -1.5 	imes 10^4 \ J$ (by convention,work done on the system is negative). Therefore,$\Delta Q = -1.5 	imes 10^4 \ J$. To convert Joules to calories,we use the conversion factor $1 \ 	ext{cal} \approx 4.18 \ J$. $\Delta Q = \frac{-1.5 	imes 10^4}{4.18} \ 	ext{cal} \approx -3.588 	imes 10^3 \ 	ext{cal} \approx -3.6 	imes 10^3 \ 	ext{cal}$. The negative sign indicates that heat flowed out of the gas.

When an ideal gas in a cylinder was compressed isothermally by a piston,the work done on the gas was found to be $1.5 \times 10^4 \ J$. During this process,about:

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