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Difficulty: Very hardPressure Law (Gay-Lussac's Law of Temperature-Pressure)

A gas sealed inside a rigid container of fixed volume is connected to a pressure gauge that displays a gauge pressure of 1.50 atm1.50\text{ atm} at an initial temperature of 27∘C27^\circ\text{C}. If the ambient atmospheric pressure is 1.00 atm1.00\text{ atm}, calculate the temperature in degrees Celsius (∘C^\circ\text{C}) to which the gas must be heated for its total absolute pressure to double.

Answer: 327 °C

Answer

The final temperature required is 327∘C327^\circ\text{C}.
To find the required temperature, first calculate the initial absolute pressure (1.50 atm+1.00 atm=2.50 atm1.50\text{ atm} + 1.00\text{ atm} = 2.50\text{ atm}) and convert the initial temperature to Kelvin (27∘C+273=300 K27^\circ\text{C} + 273 = 300\text{ K}). Doubling the absolute pressure to 5.00 atm5.00\text{ atm} requires doubling the absolute temperature according to the Pressure Law (T2=600 KT_2 = 600\text{ K}). Converting 600 K600\text{ K} back to Celsius yields 600−273=327∘C600 - 273 = 327^\circ\text{C}.

Step-by-Step Solution

1
Determine the initial absolute pressure of the gas.
P1=2.50 atmP_1 = 2.50\text{ atm}
Absolute pressure is the sum of gauge pressure and ambient atmospheric pressure (Pabs=Pgauge+PatmP_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}}).
2
Convert the initial temperature from Celsius to Kelvin.
T1=300 KT_1 = 300\text{ K}
Gas law calculations strictly require thermodynamic temperature expressed in Kelvin (TK=t∘C+273T_{\text{K}} = t_{^\circ\text{C}} + 273).
3
Calculate the required final absolute pressure.
P2=5.00 atmP_2 = 5.00\text{ atm}
The total absolute pressure doubles, so P2=2×2.50 atm=5.00 atmP_2 = 2 \times 2.50\text{ atm} = 5.00\text{ atm}.
4
Apply Gay-Lussac's Pressure Law to calculate the final absolute temperature.
T2=600 KT_2 = 600\text{ K}
At constant volume, pressure is directly proportional to absolute temperature (P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2}), yielding T2=T1×P2P1=300×2=600 KT_2 = T_1 \times \frac{P_2}{P_1} = 300 \times 2 = 600\text{ K}.
5
Convert the final absolute temperature back to degrees Celsius.
t2=327∘Ct_2 = 327^\circ\text{C}
Subtract 273 from the Kelvin temperature (t∘C=600−273=327∘Ct_{^\circ\text{C}} = 600 - 273 = 327^\circ\text{C}).

Key Concept

Pressure Law (Gay-Lussac's Law) states that the pressure of a fixed mass of gas at constant volume is directly proportional to its absolute temperature (P∝TP \propto T). Calculations must strictly use absolute pressure and Kelvin temperatures.
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Topics

Pressure Law (Gay-Lussac's Law of Temperature-Pressure)

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