Change in pressure using PV=nRT

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Homework Help Overview

The discussion revolves around a problem involving the ideal gas law (PV=nRT) where a gas is initially confined in a tank at a specific pressure and temperature. The scenario changes when half of the gas is withdrawn and the temperature is increased, prompting participants to determine the new pressure in the tank.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the ideal gas law to find the new pressure after changes in volume and temperature. They express uncertainty about their calculations and seek clarification on their reasoning.
  • Some participants question the assumption that the number of moles (n) remains constant when gas is withdrawn, suggesting that this may affect the outcome.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. There is a recognition of the need to reconsider the impact of changing the amount of gas on the calculations. No consensus has been reached yet.

Contextual Notes

The problem is constrained by the conditions of a homework assignment, which may limit the information available for discussion. Participants are navigating the implications of removing gas from the system while applying the ideal gas law.

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If half of the gas is withdrawn and the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If half of the gas is withdrawn and the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.
 
Complete Solution Removed
 
unscientific, please check your PM's.
 

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