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(D) For an ideal gas,the isothermal elasticity is equal to the pressure $P$. Thus,we need to find the ratio of the final pressure $P_2$ to the initial

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$1/3$ times

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(D) For an ideal gas,the isothermal elasticity is equal to the pressure $P$. Thus,we need to find the ratio of the final pressure $P_2$ to the initial pressure $P_1$.From the ideal gas equation,$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.Given: $V_2 = 4V_1$,$T_1 = 27 + 273 = 300 \ K$,and $T_2 = 127 + 273 = 400 \ K$.Substituting these values: $\frac{P_2}{P_1} = \frac{V_1}{V_2} 	imes \frac{T_2}{T_1} = \left( \frac{1}{4} ight) 	imes \left( \frac{400}{300} ight) = \frac{1}{4} 	imes \frac{4}{3} = \frac{1}{3}$.Therefore,the elasticity $E_2 = \frac{1}{3} E_1$,which means the elasticity becomes $1/3$ times the original value.

If the volume of a given mass of a gas is increased four times,and the temperature is raised from $27^{\circ}C$ to $127^{\circ}C$,what happens to its elasticity?

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