Pressure of a gas of constant volume at 20^{c | vedclass

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(B) Given:$T_1 = (273 + 20) \ K = 293 \ K$$p_1 = 90 \ cm 	ext{ of } Hg$$p_2 = 75 \ cm 	ext{ of } Hg$Since the volume of the gas is constant,accordin

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(B) Given:$T_1 = (273 + 20) \ K = 293 \ K$$p_1 = 90 \ cm 	ext{ of } Hg$$p_2 = 75 \ cm 	ext{ of } Hg$Since the volume of the gas is constant,according to Gay-Lussac's Law (derived from the ideal gas equation):$\frac{p_1}{T_1} = \frac{p_2}{T_2}$$\Rightarrow T_2 = \frac{T_1 	imes p_2}{p_1}$Substituting the values:$T_2 = \frac{293 	imes 75}{90} \ K$$T_2 = 244.16 \ K$To convert the temperature to Celsius:$T(^{\circ}C) = 244.16 - 273 = -28.84^{\circ} C \approx -28.8^{\circ} C$

Pressure of a gas of constant volume at $20^{\circ} C$ is $90 \ cm$ of $Hg$. At what temperature the pressure would change to $75 \ cm$ of $Hg$ (in $^{\circ} C$)?

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