In an isochoric process,if t_1 = 27^{circ}C a | vedclass

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(C) In an isochoric process,the volume of the gas remains constant.According to Gay-Lussac's law,for a fixed mass of gas at constant volume,the pressu

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$\frac{3}{4}$

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

In an isochoric process,if $t_1 = 27^{\circ}C$ and $t_2 = 127^{\circ}C$,then $\frac{P_1}{P_2}$ will be equal to [$P_1$ and $P_2$ are the pressures at $t_1^{\circ}C$ and $t_2^{\circ}C$ respectively].

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