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Zorluk: Çok zorGas Laws and the Ideal Gas Equation

A high-pressure storage vessel contains a sample of gas at an initial pressure of 2.00×105 Pa2.00 \times 10^5\text{ Pa} and a temperature of 27∘C27^\circ\text{C}. A relief valve releases one-third of the total mass of the gas while maintaining a constant internal volume. After the valve closes, the vessel and remaining gas are heated to 127∘C127^\circ\text{C}. What is the final pressure of the gas inside the vessel?

  1. 1.78×105 Pa1.78 \times 10^5\text{ Pa}Cevap
  2. B
    6.27×105 Pa6.27 \times 10^5\text{ Pa}
  3. C
    8.89×104 Pa8.89 \times 10^4\text{ Pa}
  4. D
    2.67×105 Pa2.67 \times 10^5\text{ Pa}

Cevap

The final pressure of the gas inside the vessel is 1.78×105 Pa1.78 \times 10^5\text{ Pa}.
According to the ideal gas equation PV=mMRTPV = \frac{m}{M}RT, pressure is directly proportional to mass and absolute temperature at fixed volume (P∝mTP \propto m T). Converting temperatures to Kelvin gives T1=300 KT_1 = 300\text{ K} and T2=400 KT_2 = 400\text{ K}. Since one-third of the gas escaped, two-thirds remains (m2=23m1m_2 = \frac{2}{3}m_1). The final pressure is therefore P2=P1×23×400300=2.00×105 Pa×89≈1.78×105 PaP_2 = P_1 \times \frac{2}{3} \times \frac{400}{300} = 2.00 \times 10^5\text{ Pa} \times \frac{8}{9} \approx 1.78 \times 10^5\text{ Pa}.

Adım Adım Çözüm

1
Convert given initial and final temperatures from Celsius to Kelvin.
T1=27∘C+273=300 KT_1 = 27^\circ\text{C} + 273 = 300\text{ K} and T2=127∘C+273=400 KT_2 = 127^\circ\text{C} + 273 = 400\text{ K}.
Absolute temperature in Kelvin is required for all gas law calculations.
2
Determine the remaining mass fraction of the gas.
m2=m1−13m1=23m1m_2 = m_1 - \frac{1}{3}m_1 = \frac{2}{3}m_1.
The pressure depends on the quantity of gas remaining in the rigid container after venting.
3
Apply the ideal gas equation PV=nRT=mMRTPV = nRT = \frac{m}{M}RT for constant volume VV and molar mass MM.
P2P1=(m2m1)(T2T1)\frac{P_2}{P_1} = \left(\frac{m_2}{m_1}\right) \left(\frac{T_2}{T_1}\right).
Pressure is directly proportional to both mass and absolute temperature when volume is constant.
4
Substitute the known values to compute P2P_2.
P2=(2.00×105 Pa)×(23)×(400 K300 K)=2.00×105×89≈1.78×105 PaP_2 = (2.00 \times 10^5\text{ Pa}) \times \left(\frac{2}{3}\right) \times \left(\frac{400\text{ K}}{300\text{ K}}\right) = 2.00 \times 10^5 \times \frac{8}{9} \approx 1.78 \times 10^5\text{ Pa}.
Evaluates the combined effects of mass reduction and temperature elevation.

Anahtar Kavram

Ideal Gas Law variations involving changing gas mass and absolute temperature at constant volume.
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Gas Laws and the Ideal Gas Equation

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