Change in pressure using PV=nRT

AI Thread Summary
The discussion revolves around calculating the new pressure of a gas after half of it is withdrawn and the temperature is increased. Initially, the gas is at 6.5 atm and 9.4°C, which converts to 282.4 K. After removing half the gas and raising the temperature to 62.4°C (335.4 K), the user attempts to apply the ideal gas law but realizes that the number of moles (n) is not constant due to the gas withdrawal. The user initially calculates a pressure of 15.439 atm but acknowledges this result is incorrect. The key issue identified is the change in the number of moles of gas, which must be accounted for in the calculations.
rinarez7
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1. Gas is confined in a tank at a pressure of
6.5 atm and a temperature of 9.4◦C.
If half of the gas is withdrawn and the
temperature is raised to 62.4◦C, what is the
new pressure in the tank? Answer in units of
atm.



2. PV=nRT
Ti= 9.4 C + 273= 282.4K
Tf= 62.4 C +273= 335.4K




3. PV= nRT, n and R are constants, therefore I used PV= T
So, initially, 6.5atmVi= 282.4K giving, Vi = 43.446

I then used (1/2) 43.446 as Vf and solved for Pf
Pf= Tf/ Vf=> 15.439 atm

But this isn't right, I know. What am I missing??
 
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Hi rinarez7,

rinarez7 said:
1. Gas is confined in a tank at a pressure of
6.5 atm and a temperature of 9.4◦C.
If half of the gas is withdrawn and the
temperature is raised to 62.4◦C, what is the
new pressure in the tank? Answer in units of
atm.



2. PV=nRT
Ti= 9.4 C + 273= 282.4K
Tf= 62.4 C +273= 335.4K




3. PV= nRT, n and R are constants


I don't believe n is a constant here; they are removing some of the gas.
 
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unscientific, please check your PM's.
 
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