Pressure-temperature Law

Question

A rigid sealed cylinder holds a gas at 2.0 	imes 10^5 Pa and 300 K; find the new pressure when the temperature is raised to 450 K at constant volume.

RankGuru · Accepted Answer

At constant volume the pressure of a fixed amount of ideal gas is directly proportional to its absolute temperature, a result known as Gay-Lussac's law: $\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$. This follows from $PV=nRT$ with $V$ and $n$ held fixed, and microscopically it reflects that hotter molecules strike the walls both harder and more frequently. Solving for the new pressure, $P_2=P_1\dfrac{T_2}{T_1}=2.0	imes10^{5}	imes\dfrac{450}{300}=3.0	imes10^{5}$ Pa. The temperatures must be expressed in kelvin for the proportionality to hold correctly. The value $1.33	imes10^{5}$ Pa inverts the temperature ratio. The value $4.5	imes10^{5}$ Pa mistakenly multiplies by an incorrect ratio arising from a celsius-to-kelvin confusion. The value $2.5	imes10^{5}$ Pa wrongly adds a fixed increment rather than scaling proportionally. This is the NCERT pressure–temperature law for an isochoric process, derived directly from the ideal gas equation and important for understanding why sealed containers can burst when heated near a flame and why pressure cookers and aerosol cans carry temperature warnings. As a plausibility check, raising the temperature by 50 % (from 300 K to 450 K) should raise the pressure by the same 50 %, taking $2.0	imes10^{5}$ Pa to $3.0	imes10^{5}$ Pa, exactly matching our computed result.

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