# Ideal gas question on glass bulbs

1. Aug 1, 2009

### fluppocinonys

Two glass bulbs of equal volume are joined by a narrow tube and are filled with gas at s.t.p. (standard temperature and pressure where $$\theta = {0^ \circ }{\rm{C}}$$ and $$p = 1.01 \times {10^5}$$ Pa). When one bulb is kept in melting ice and the other is placed in hot bath, the new pressure is 1.166 x 105 Pa. Calculate the temperature of the bath.

I tried to use
$$\begin{array}{l} \frac{{{p_1}{v_1}}}{{{T_1}}} = \frac{{{p_2}{v_2}}}{{{T_2}}} \\ {v_1} = {v_2} \\ \frac{{{p_1}}}{{{T_1}}} = \frac{{{p_2}}}{{{T_2}}} \\ \end{array}$$

Can anyone guide me? Thanks.

2. Aug 1, 2009

### kuruman

Careful here. When one bulb is placed in a hot bath, is its temperature the only thing that changes? Hint: The bulb that is at 0 oC has its final temperature and volume the same, but its pressure increases. How can that be?

3. Aug 1, 2009

### fluppocinonys

The bulb at 0C has higher pressure because amount of gas molecules are higher at there?
But how could it be since it has lower temperature, i thought the pressure would be higher in the bulb at hot bath since the gas molecules travel faster

4. Aug 1, 2009

### kuruman

No. Remember that the pressure is the same in both bulbs. Even though when you heat up the molecules and they move faster (on the average) there are fewer of them. When you raise the temperature of a gas, its density decreases as molecules leave the gas (if they are allowed to do so). That's how hot air balloons float in the (relatively) cooler air that surrounds them.

5. Aug 1, 2009

### fluppocinonys

ok, the bulb placed in hot bath will increase its temperature and pressure, that's why the overall pressure is increased, wait, that means
$$\begin{array}{l} \frac{{{p_1}{v_1}}}{{{T_1}}} = \frac{{{p_2}{v_2}}}{{{T_2}}} \\ \end{array}$$
is not valid in this situation?

6. Aug 1, 2009

### kuruman

Yes, it is not valid. Initially, there are equal numbers of molecules in each bulb. When the temperature is increased in one bulb, molecules leave it and go into the other bulb. You need to take that into account.

7. Aug 1, 2009

### fluppocinonys

so is it solve by this way ?

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8. Aug 1, 2009

### kuruman

Basically, correct. If you want to be 100% correct, you should multiply the right side of the first two equations by $$\frac{V}{R}$$. When you put the two together as in the last equation, the ratio will appear on both sides and will drop out.

9. Aug 1, 2009

### fluppocinonys

All right thank you very much!