Really simple ideal gas question.

In summary, the question involves a gas in a tank at a pressure of 10 atm and a temperature of 15C. After half of the gas is withdrawn and the temperature is raised to 65C, the new pressure needs to be calculated. Using the equation pV=nRT, the solution involves converting the pressure to kPa and using the constant R = 0.08. The final answer is 5.87 atm. It is important to always include unit conversions to ensure correct calculations.
  • #1
kidwithshirt
11
0
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gas is confined in a tank at pressure of 10 atm and temp of 15C, if half of the gas is withdrawn and temp is raised to 65C, what's the new pressure?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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  • #2
P1V = nRT1 and P2V = (n/2)RT2. Does this help?
 
  • #3
i keep getting 21.6.

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

answer is around 5.7
 
  • #4
I get 5.87 atm. Divide 2nd eqn by 1st.
 
  • #5
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.
 
  • #6
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?
 
  • #7
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
 
  • #8
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)
 

FAQ: Really simple ideal gas question.

What is an ideal gas?

An ideal gas is a theoretical concept in which the gas particles have no volume and do not interact with each other. This means that the gas follows the ideal gas law, which states that the pressure, volume, and temperature of the gas are all related.

What is the ideal gas law?

The ideal gas law is a mathematical equation that describes the relationship between the pressure, volume, and temperature of an ideal gas. It is written as PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature.

How do you calculate the pressure of an ideal gas?

The pressure of an ideal gas can be calculated using the ideal gas law, PV = nRT. Simply rearrange the equation to solve for P, and plug in the values for the other variables.

What is the relationship between temperature and pressure in an ideal gas?

According to the ideal gas law, the temperature and pressure of an ideal gas are directly proportional. This means that as the temperature of the gas increases, so does the pressure, and vice versa.

Why is the ideal gas law only applicable to ideal gases?

The ideal gas law is only applicable to ideal gases because real gases do have volume and do interact with each other. Therefore, the behavior of real gases cannot be accurately described using the ideal gas law. However, the ideal gas law can be a good approximation for real gases under certain conditions, such as high temperature and low pressure.

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