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Zorluk: ZorGas Laws and the Ideal Gas Equation

A rigid gas vessel sealed on a meteorological satellite initially contains a fixed volume of an ideal gas at a pressure of 1.50×105 Pa1.50 \times 10^5\text{ Pa} and a temperature of 27C27^\circ\text{C}. During an orbital maneuver, solar radiation heats the vessel to 327C327^\circ\text{C}, while a small vent valve simultaneously allows 20%20\% of the gas molecules to escape. Assuming the volume of the vessel remains constant, what is the final pressure of the gas inside the vessel?

  1. 2.40×105 Pa2.40 \times 10^5\text{ Pa}Cevap
  2. B
    1.45×106 Pa1.45 \times 10^6\text{ Pa}
  3. C
    3.00×105 Pa3.00 \times 10^5\text{ Pa}
  4. D
    6.00×104 Pa6.00 \times 10^4\text{ Pa}

Cevap

2.40×105 Pa2.40 \times 10^5\text{ Pa}
According to the ideal gas law (PV=nRTPV = nRT), when volume is held constant, pressure is directly proportional to both the number of moles of gas present and the absolute temperature in Kelvin (PnTP \propto nT). First, convert the initial and final temperatures to Kelvin: T1=27+273=300 KT_1 = 27 + 273 = 300\text{ K} and T2=327+273=600 KT_2 = 327 + 273 = 600\text{ K}. Because 20%20\% of the gas escapes, 80%80\% remains inside the vessel, so n2=0.80n1n_2 = 0.80n_1. Calculating final pressure yields P2=P1×(n2/n1)×(T2/T1)=1.50×105×0.80×(600/300)=2.40×105 PaP_2 = P_1 \times (n_2/n_1) \times (T_2/T_1) = 1.50 \times 10^5 \times 0.80 \times (600/300) = 2.40 \times 10^5\text{ Pa}.

Adım Adım Çözüm

1
Convert all given temperatures from degrees Celsius to absolute temperature in Kelvin.
T1=27+273.15=300 KT_1 = 27 + 273.15 = 300\text{ K}, T2=327+273.15=600 KT_2 = 327 + 273.15 = 600\text{ K}
Gas laws and the ideal gas equation require absolute temperature in Kelvin.
2
Determine the fraction of gas remaining in the rigid vessel.
If 20%20\% escapes, remaining fraction n2=(10.20)n1=0.80n1n_2 = (1 - 0.20)n_1 = 0.80n_1
Pressure depends on the quantity of gas remaining inside the fixed volume.
3
Apply the ideal gas equation PV=nRTPV = nRT for constant volume VV.
P2=P1×(n2n1)×(T2T1)P_2 = P_1 \times \left(\frac{n_2}{n_1}\right) \times \left(\frac{T_2}{T_1}\right)
Since volume is constant, pressure is directly proportional to the product of amount of gas and absolute temperature.
4
Substitute known values to compute the final pressure P2P_2.
P2=(1.50×105 Pa)×0.80×(600 K300 K)=2.40×105 PaP_2 = (1.50 \times 10^5\text{ Pa}) \times 0.80 \times \left(\frac{600\text{ K}}{300\text{ K}}\right) = 2.40 \times 10^5\text{ Pa}
Multiplying the initial pressure by the mole fraction ratio and temperature ratio gives the final gas pressure.

Anahtar Kavram

Ideal Gas Equation and Combined Gas Behavior
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