Change in pressure using PV=nRT

Click For Summary

Homework Help Overview

The problem involves a gas confined in a tank, initially at a pressure of 6.5 atm and a temperature of 9.4°C. The scenario changes when half of the gas is withdrawn and the temperature is increased to 62.4°C, prompting a question about the new pressure in the tank.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of withdrawing half the gas, questioning whether this affects the number of particles (n) or the volume (V) of the gas. There is an exploration of the application of the ideal gas law, PV=nRT, and how to correctly interpret the changes in conditions.

Discussion Status

Participants are actively engaging with the problem, offering different interpretations of the effects of withdrawing gas on the pressure calculation. Some guidance has been provided regarding the treatment of the number of particles and the volume, but no consensus has been reached on the correct approach.

Contextual Notes

The problem does not specify changes to the volume of the tank, leading to differing assumptions about how to apply the ideal gas law in this context.

rinarez7
Messages
27
Reaction score
0
1b]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If half of the gas is withdrawn and the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?? .
 
Physics news on Phys.org
My thought would be that "half of the gas is withdrawn" means that the number of particles n is halved, and the volume is kept constant, not the other way around.
 
It doesn't mention the volume has been reduced by a half, the amount of gas has been reduced by a half so you need to use 0.5n.
 
Thanks to the both of you!
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
9K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
9K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
1
Views
2K