Really simple ideal gas question.

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Homework Help Overview

The problem involves an ideal gas scenario where the gas is initially at a pressure of 10 atm and a temperature of 15°C. After withdrawing half of the gas and raising the temperature to 65°C, the question seeks to determine the new pressure.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the ideal gas law, with some attempting to relate initial and final states using equations. There are questions about unit conversions and the implications of temperature changes on pressure calculations.

Discussion Status

Several participants have shared their calculations, with some arriving at different pressure values. There is an ongoing exploration of the correct approach, and guidance regarding unit conversions has been offered. The discussion reflects a mix of interpretations and attempts to clarify the problem.

Contextual Notes

There is mention of potential confusion regarding temperature units and the need for conversions, as well as the impact of using different values for the gas constant. Participants express frustration over the problem's complexity and the time spent on it.

kidwithshirt
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You know what, I think I might be stupid. I tried this question for 20 minutes already and I am still stucked

Homework Statement


gas is confined in a tank at pressure of 10 atm and temp of 15C, if half of the gas is withdrawn and temp is raised to 65C, what's the new pressure?


Homework Equations


pV=nRT


The Attempt at a Solution



Don't even joke about this.

Help would be welcomed.
 
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P1V = nRT1 and P2V = (n/2)RT2. Does this help?
 
i keep getting 21.6.

does it reduce to T1/P1=T2/(2*P2)?

answer is around 5.7
 
I get 5.87 atm. Divide 2nd eqn by 1st.
 
can you show me the steps.

i spent at least two hours on this already.

After going through two years of calculus I can't even do this stupid problem.
 
I'm not sure if this helps but if you use the pv = nRT formula, your pressure would have to be in kPa. Have you converted your value of atm from the given question?
 
Shooting star said:
I get 5.87 atm. Divide 2nd eqn by 1st.

never mind. I've been doing it right since the beginning

I just forgot the it's in K.

I thought using the 0.08 for R can allow you to omit the C -> K conversion

That's why I've been getting 21.6 all the time

thank you anyways
 
fyi, always use the units, even when bringing in a constant equation like that. In first year my chem prof use to take marks off for not including the unit conversions. This way you know that when the units cross out, its done correctly (units wise)
 

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