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Zorluk: Çok zorGas Laws and the Ideal Gas Equation

A meteorological balloon filled with 0.50 kg0.50\text{ kg} of helium gas is launched at sea level, where the atmospheric pressure is 1.01×105 Pa1.01 \times 10^5\text{ Pa} and the ambient temperature is 27C27^\circ\text{C}. The balloon ascends to a high altitude where the external ambient pressure decreases to 4.00×104 Pa4.00 \times 10^4\text{ Pa} and the ambient temperature drops to 23C-23^\circ\text{C}. As the balloon expands, its elastic membrane exerts an additional pressure, causing the internal gas pressure to be 20%20\% higher than the surrounding ambient pressure. Assuming helium behaves as an ideal gas with a molar mass of 4.0 g/mol4.0\text{ g/mol} and the molar gas constant R=8.31 J mol1K1R = 8.31\text{ J mol}^{-1}\text{K}^{-1}, calculate the final volume of helium gas inside the balloon at this altitude in m3\text{m}^3.

Cevap: 5.41 m^3

Cevap

The final volume of helium gas inside the balloon at altitude is 5.41 m35.41\text{ m}^3.
The ideal gas equation PV=nRTPV = nRT relates state variables. By determining n=125 molesn = 125\text{ moles} from mass and molar mass, absolute temperature T2=250 KT_2 = 250\text{ K}, and total internal pressure P2=4.80×104 PaP_2 = 4.80 \times 10^4\text{ Pa}, the final volume V2V_2 evaluates to 5.41 m35.41\text{ m}^3.

Adım Adım Çözüm

1
Convert the mass of helium into moles
n=125 molesn = 125\text{ moles}
Mass m=0.50 kg=500 gm = 0.50\text{ kg} = 500\text{ g} divided by molar mass M=4.0 g/molM = 4.0\text{ g/mol} gives n=5004.0=125 molesn = \frac{500}{4.0} = 125\text{ moles}.
2
Convert final temperature to Kelvin
T2=250 KT_2 = 250\text{ K}
Gas laws strictly require thermodynamic temperature: T2=23+273=250 KT_2 = -23 + 273 = 250\text{ K}.
3
Calculate final internal pressure of the gas
P2=4.80×104 PaP_2 = 4.80 \times 10^4\text{ Pa}
The gas pressure inside the balloon is 20%20\% higher than external ambient pressure: P2=1.20×(4.00×104)=4.80×104 PaP_2 = 1.20 \times (4.00 \times 10^4) = 4.80 \times 10^4\text{ Pa}.
4
Solve for the final volume using the ideal gas equation
V2=5.41 m3V_2 = 5.41\text{ m}^3
Rearranging P2V2=nRT2P_2 V_2 = n R T_2 yields V2=nRT2P2=125×8.31×2504.80×104=5.41015... m35.41 m3V_2 = \frac{n R T_2}{P_2} = \frac{125 \times 8.31 \times 250}{4.80 \times 10^4} = 5.41015...\text{ m}^3 \approx 5.41\text{ m}^3.

Anahtar Kavram

Ideal Gas Equation (PV=nRTPV = nRT)
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