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Difficulty: Very hardGas Laws and the Ideal Gas Equation

A gas cylinder fitted with a pressure relief valve contains a fixed mass of an ideal gas at an initial absolute pressure of 2.50×105 Pa2.50 \times 10^5\text{ Pa} and a temperature of 27∘C27^\circ\text{C}. The valve is designed to open when the internal gauge pressure exceeds 3.20×105 Pa3.20 \times 10^5\text{ Pa}. Assuming the atmospheric pressure is 1.00×105 Pa1.00 \times 10^5\text{ Pa}, at what temperature will the relief valve open?

  1. 231∘C231^\circ\text{C}Answer
  2. B
    45.4∘C45.4^\circ\text{C}
  3. C
    111∘C111^\circ\text{C}
  4. D
    34.6∘C34.6^\circ\text{C}

Answer

231∘C231^\circ\text{C}
The correct answer is 231∘C231^\circ\text{C}. To solve this, first convert the initial temperature to Kelvin: T1=27+273=300 KT_1 = 27 + 273 = 300\text{ K}. Next, calculate the final absolute pressure by adding the atmospheric pressure to the given gauge pressure: P2=3.20×105 Pa+1.00×105 Pa=4.20×105 PaP_2 = 3.20 \times 10^5\text{ Pa} + 1.00 \times 10^5\text{ Pa} = 4.20 \times 10^5\text{ Pa}. Applying Gay-Lussac's Law (P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2}) gives T2=300×4.20×1052.50×105=504 KT_2 = 300 \times \frac{4.20 \times 10^5}{2.50 \times 10^5} = 504\text{ K}. Converting back to Celsius gives 504−273=231∘C504 - 273 = 231^\circ\text{C}.

Step-by-Step Solution

1
Convert the initial temperature from Celsius to Kelvin and determine the initial absolute pressure.
T1=27+273=300 KT_1 = 27 + 273 = 300\text{ K}, P1=2.50×105 PaP_1 = 2.50 \times 10^5\text{ Pa}.
Gas laws require thermodynamic (absolute) temperature in Kelvin.
2
Calculate the final total absolute pressure (P2P_2) at which the valve opens.
P2=Pgauge+Patm=3.20×105 Pa+1.00×105 Pa=4.20×105 PaP_2 = P_{\text{gauge}} + P_{\text{atm}} = 3.20 \times 10^5\text{ Pa} + 1.00 \times 10^5\text{ Pa} = 4.20 \times 10^5\text{ Pa}.
Gauge pressure measures the pressure difference relative to atmospheric pressure; gas laws use absolute pressure.
3
Apply Gay-Lussac's (Pressure) Law at constant volume to find the final Kelvin temperature (T2T_2).
P1T1=P2T2  ⟹  T2=300×4.20×1052.50×105=504 K\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies T_2 = 300 \times \frac{4.20 \times 10^5}{2.50 \times 10^5} = 504\text{ K}.
For a fixed volume of ideal gas, absolute pressure is directly proportional to absolute temperature.
4
Convert the final temperature back to degrees Celsius.
t2=504−273=231∘Ct_2 = 504 - 273 = 231^\circ\text{C}.
The question asks for the temperature in degrees Celsius.

Key Concept

Pressure Law (Gay-Lussac's Law) and Absolute vs Gauge Pressure
Estimated Time:2m 0s
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Topics

Gas Laws and the Ideal Gas Equation

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