10 moles of an ideal gas has a gauge pressure of 2 atm what is new T? no clue

[teggenspiller]

10 moles of an ideal gas has a gauge pressure of 2 atm what is new T?? no clue

1. Homework Statement [/b
10 moles of an ideal gas has a gauge pressure of 2 atm and a temperature of 200 K. If the volume of the gas is doubled and the pressure dropped to 1 atm., what is the new temperature?

Homework Equations

i have no clue how to do this.
i know that PV=nRT= constant. but when i try to sub things out or solve for V first, i get so lost.
ive been working on this for hours.

So can someone just please show me how to do this? step by step?

 

[zhermes]

your equation isn't quite right.
PV = nRT yes. But each side is not necessarily constant. The 'R' in the equation, is necessarily constant (it is defined as a constant). Therefore:
\frac{PV}{nT} = R = constant
This means that if we go from starting conditions ('1') to ending conditions ('2'):
\frac{P_1 V_1}{n_1 T_1} = \frac{P_2 V_2}{n_2 T_2}
That's the equation you always want to start with. You just need to start plugging in.

Note that in your case, the number of moles (n) stays the same, therefore n_1 = n_2 and they cancel out of the equation. Anytime a quantity stays the same, it will cancel out. Plug in the values for what you know, and solve for what you don't.

 

[Yuqing]

PV = nRT does not equal a constant. The only known constants in the ideal gas equation currently are the ideal gas constant R and the number of moles n. You can safely write that
PV/T = nR = constant and use this to connect the two equations.

 

[teggenspiller]

ha! thank you 2 so very much. May I put in "2v" in place of the second V? It's value isn't nessecarily important, right?

 

[teggenspiller]

Tf= 10 * (202650* V/10*200) / 1* 2V
??

 

[Yuqing]

You should be able to cancel the volumes.

 

[gneill]

ha! thank you 2 so very much. May I put in "2v" in place of the second V? It's value isn't nessecarily important, right?

Sure. And put in p1/2 for p2. You can solve this problem symbolically without plugging in any numbers!

 

[teggenspiller]

i think may equal 400 K

 

[gneill]

i think may equal 400 K

And the difference between thinking and knowing is...?

Just look at the equation:

PV = nRT

You're doubling V and halving P. What's the net change to PV?

 

[teggenspiller]

well, it seems that PV would be the same as 2P1/2V
because there values would even out? so temp is related to that how.. well if we arent including n and R. and 200= PV and T2 = 1/2P* 2V himm... why don't i get it?

 

[teggenspiller]

200k = 200k duh. tis 200

 

[gneill]

200k = 200k duh. tis 200

Yes! I new you'd get there in the end.

 

[teggenspiller]

its amazing when you come back to something with a clear head.

 

[ehild]

Take care. It is said that the gauge pressure was 2 atm.
The gauge pressure is (cited from http://www.britannica.com/)

"The reading on a gauge, which is the difference between two pressures, is known as the gauge pressure. If the lower of the pressures is the pressure of the atmosphere, the total, or absolute, pressure is the sum of the gauge and atmospheric pressures."

2 atm gauge pressure corresponds to 3 atm absolute pressure.You have to use absolute pressures in the ideal gas law.

ehild

 

[gneill]

Take care. It is said that the gauge pressure was 2 atm.
The gauge pressure is (cited from http://www.britannica.com/)

"The reading on a gauge, which is the difference between two pressures, is known as the gauge pressure. If the lower of the pressures is the pressure of the atmosphere, the total, or absolute, pressure is the sum of the gauge and atmospheric pressures."

2 atm gauge pressure corresponds to 3 atm absolute pressure.You have to use absolute pressures in the ideal gas law.

ehild

D'oh! Good catch ehild! Don't know why I missed that.

teggenspiller, the absolute pressure is not being halved. The gauge pressure is going from 2 atm to 1 atm. What does that imply about the pressure ratio?