What is the resultant pressure if 1.7 mol of ideal gas at 273 K and 2.79 atm in a closed container of constant volume is heated
1 answer:
Answer: The resultant pressure is 3.22 atm
Explanation:
Gay-Lussac's Law: This law states that pressure is directly proportional to the temperature of the gas at constant volume and number of moles.
(At constant volume and number of moles)
where,
= initial pressure of gas = 2.79 atm
= final pressure of gas = ?
= initial temperature of gas = 273K
= final temperature of gas = 315 K
Thus the resultant pressure is 3.22 atm
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Answer : The final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.
Explanation :
As we know that:
At constant volume and temperature of gas, the pressure will be directly proportional to the number of moles of gas.
The relation between pressure and number of moles of gas will be:
where,
= initial pressure of gas = 24.5 atm
= final pressure of gas = 5.30 atm
= initial number of moles of gas = 1.40 moles
= final number of moles of gas = ?
Now put all the given values in the above expression, we get:
Therefore, the final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.