गे-लुसैक का नियम (स्थिर आयतन) – समस्याएं और समाधान

1. Ideal gases initially have दबाव P and तापमान T. The gas undergoes the आइसोकोरिक प्रक्रिया so that the final pressure becomes 4 times the initial pressure. What is the final temperature of the gas?

ज्ञात :

प्रारंभिक दबाव (P1) = पी

अंतिम दबाव (पी)2) = 4पी

प्रारंभिक तापमान (T)1) = टी

चाहता था: अंतिम तापमान (T)2)

समाधान:

का सूत्र गे-लुसाक का नियम :

Gay-Lussac's law (constant volume) - problems and solutions 1

The final temperature becomes 4 times the initial temperature.

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2. In a closed container, ideal gases initially have a temperature of 27oC. If the final pressure becomes 2 times the initial pressure, what is the final temperature?

ज्ञात :

प्रारंभिक दबाव (P1) = पी

अंतिम दबाव (पी)2) = 2पी

प्रारंभिक तापमान (T)1) = 27oC + 273 = 300 K

चाहता था: अंतिम तापमान (T)2)

समाधान:

Gay-Lussac's law (constant volume) - problems and solutions 2

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3. A tire is filled to a gauge pressure of 2 atm at 27°C. After a drive, the temperature within the tire rises to 47°C. What is the pressure within the tire now?

ज्ञात :

The atmospheric pressure = 1 atm = 1 x 105 Pa

The initial gauge pressure = 2 atm = 2 x 105 Pa

The initial absolute pressure (P1) = 1 atm + 2 atm = 3 atm = 3 x 105 Pa

The initial temperature (T1) = 27oC + 273 = 300 K

The final temperature (T1) = 47oC + 273 = 320 K

वांछित : The final gauge temperature

समाधान:

Gay-Lussac's law (constant volume) - problems and solutions 3

The final gauge pressure = final absolute pressure – atmospheric pressure

The final gauge pressure = 3.2 atm – 1 atm

The final gauge pressure = 2.2 atm

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