In an isochoric process,if {T_1} = {27^o}C an | vedclass

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(D) In an isochoric process,the volume $V$ remains constant.According to Gay-Lussac's Law,for a fixed mass of an ideal gas at constant volume,the pres

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$3/4$

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(D) In an isochoric process,the volume $V$ remains constant.According to Gay-Lussac's Law,for a fixed mass of an ideal gas at constant volume,the pressure $P$ is directly proportional to the absolute temperature $T$ $(P \propto T)$.Therefore,$\frac{P_1}{P_2} = \frac{T_1}{T_2}$.First,convert the temperatures from Celsius to Kelvin:$T_1 = 27^o C = 27 + 273 = 300 \ K$$T_2 = 127^o C = 127 + 273 = 400 \ K$Now,substitute these values into the ratio:$\frac{P_1}{P_2} = \frac{300}{400} = \frac{3}{4}$.

In an isochoric process,if ${T_1} = {27^o}C$ and ${T_2} = {127^o}C$,then ${P_1}/{P_2}$ will be equal to:

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