Really simple ideal gas question.

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The discussion revolves around a gas problem involving pressure, temperature, and the ideal gas law. A user struggles to find the new pressure after withdrawing half the gas and increasing the temperature, initially calculating incorrect values. After multiple attempts, it is revealed that the failure to convert Celsius to Kelvin led to persistent errors. The correct answer is approximately 5.87 atm, emphasizing the importance of unit conversions in calculations. The conversation highlights the significance of careful unit management in solving gas law problems.
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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gas is confined in a tank at pressure of 10 atm and temp of 15C, if half of the gas is withdrawn and temp is raised to 65C, what's the new pressure?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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