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(A) For an isobaric process,the heat added is given by $E_1 = n C_p \Delta T$. The change in internal energy is given by $E_2 = n C_v \Delta T$. There

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$15$

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(A) For an isobaric process,the heat added is given by $E_1 = n C_p \Delta T$. The change in internal energy is given by $E_2 = n C_v \Delta T$. Therefore,the ratio is $\frac{E_1}{E_2} = \frac{n C_p \Delta T}{n C_v \Delta T} = \frac{C_p}{C_v} = \gamma$. For a monoatomic gas,the degrees of freedom $f = 3$. The adiabatic index is $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$. Given $\frac{E_1}{E_2} = \frac{x}{9}$,we have $\frac{5}{3} = \frac{x}{9}$. Solving for $x$,we get $x = \frac{5 	imes 9}{3} = 15$.

The temperature of $1$ mole of an ideal monoatomic gas is increased by $50^{\circ} C$ at constant pressure. The total heat added and change in internal energy are $E_1$ and $E_2$,respectively. If $\frac{E_1}{E_2}=\frac{x}{9}$,then the value of $x$ is . . . . . . .

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