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Zorluk: ZorPressure Law (Gay-Lussac's Law of Temperature-Pressure)

A sample of nitrogen gas enclosed in a constant-volume container exerts a pressure of 2.50 atm2.50\text{ atm} at a temperature of 23C-23^\circ\text{C}. To what temperature, in degrees Celsius (C^\circ\text{C}), must the gas be heated so that its pressure increases to 4.00 atm4.00\text{ atm}?

Cevap: 127 °C

Cevap

The gas must be heated to 127C127^\circ\text{C}.
According to Gay-Lussac's Pressure Law, for a fixed mass of gas at constant volume, pressure is directly proportional to absolute temperature (PTP \propto T). Converting 23C-23^\circ\text{C} to Kelvin gives 250 K250\text{ K}. Solving 2.50250=4.00T2\frac{2.50}{250} = \frac{4.00}{T_2} gives T2=400 KT_2 = 400\text{ K}. Converting back to Celsius (400273400 - 273) yields the correct temperature of 127C127^\circ\text{C}.

Adım Adım Çözüm

1
Convert the initial temperature from Celsius to the thermodynamic temperature scale (Kelvin).
T1=23+273=250 KT_1 = -23 + 273 = 250\text{ K}
Gas laws strictly require temperature to be expressed in Kelvin.
2
Use Gay-Lussac's Pressure Law equation P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2} to solve for T2T_2.
T2=P2×T1P1=4.00×2502.50=400 KT_2 = \frac{P_2 \times T_1}{P_1} = \frac{4.00 \times 250}{2.50} = 400\text{ K}
At constant volume, the pressure of a given mass of gas is directly proportional to its absolute temperature.
3
Convert the calculated temperature T2T_2 back to degrees Celsius.
t2=400273=127Ct_2 = 400 - 273 = 127^\circ\text{C}
The question explicitly requests the final temperature in degrees Celsius.

Anahtar Kavram

Pressure Law (Gay-Lussac's Law)
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