- #1
plain stupid
- 18
- 1
- Homework Statement
- At the top of the sealed off end of a mercury manometer U-tube there's a gas: at 30 degrees Celsius, its volume is 50 cubic centimeters. The mercury level is 10 cm lower in the closed end than in the open end of the tube. The barometer shows a pressure of 75 mmHg. Find the normal volume of the gas.
- Relevant Equations
- pV = nRT,
p = ρgh
Some notation:
- the difference between the heights of mercury, which is effectively the height of the mercury in the open end of the tube is ##h_{diff}##
- the volume of gas inside the sealed off end is ##V_{inside}##
- the volume of gas when let outside, "normal volume", is ##V_{outside}##
- the amount of the gas in moles is ##n##
- the temperature of gas is ##T##
- the pressure measured by the barometer is ##p_{outside}##
- the density of mercury is ##\rho##
- the mercury level in the barometer (measuring outside pressure, I assume) is ##h_{barometer}##
The pressure of the gas in the sealed off end should be balanced by the pressure of the mercury level difference in the open end + the pressure the barometer's showing (outside pressure):
$$\frac{nRT}{V_{inside}} = \rho g \left(h_{diff}+h_{barometer}\right)$$
From this I get the amount of substance ##n## as
$$n = \frac{\rho g \left(h_{diff}+h_{barometer}\right) V_{inside} }{RT}$$
I'm not sure what "normal volume" is, but I assume it just means outside of the tube, at the same temperature, i.e. ##T = 30^\circ C##:
$$V_{outside} = \frac{nRT}{p_{outside}} = \frac{\frac{\rho g \left(h_{diff}+h_{barometer}\right) V_{inside} }{RT} \cdot RT}{\rho g h_{barometer}} $$
$$V_{outside} = \frac{\left(h_{diff}+h_{barometer}\right) V_{inside}}{h_{barometer}}$$
This gives me the wrong result. The correct result is around ##50.4 ~ cm^3##, and I get ##56.67##, so... I'm missing something again. I think my main idea is correct (the first two equations), but at the end I assume that in order to get this "normal" volume of the gas, it should be outside the tube, and its pressure should be equal to the outside pressure, so that's why I plug that into the formula.
I'd appreciate any hints you might throw at me.
- the difference between the heights of mercury, which is effectively the height of the mercury in the open end of the tube is ##h_{diff}##
- the volume of gas inside the sealed off end is ##V_{inside}##
- the volume of gas when let outside, "normal volume", is ##V_{outside}##
- the amount of the gas in moles is ##n##
- the temperature of gas is ##T##
- the pressure measured by the barometer is ##p_{outside}##
- the density of mercury is ##\rho##
- the mercury level in the barometer (measuring outside pressure, I assume) is ##h_{barometer}##
The pressure of the gas in the sealed off end should be balanced by the pressure of the mercury level difference in the open end + the pressure the barometer's showing (outside pressure):
$$\frac{nRT}{V_{inside}} = \rho g \left(h_{diff}+h_{barometer}\right)$$
From this I get the amount of substance ##n## as
$$n = \frac{\rho g \left(h_{diff}+h_{barometer}\right) V_{inside} }{RT}$$
I'm not sure what "normal volume" is, but I assume it just means outside of the tube, at the same temperature, i.e. ##T = 30^\circ C##:
$$V_{outside} = \frac{nRT}{p_{outside}} = \frac{\frac{\rho g \left(h_{diff}+h_{barometer}\right) V_{inside} }{RT} \cdot RT}{\rho g h_{barometer}} $$
$$V_{outside} = \frac{\left(h_{diff}+h_{barometer}\right) V_{inside}}{h_{barometer}}$$
This gives me the wrong result. The correct result is around ##50.4 ~ cm^3##, and I get ##56.67##, so... I'm missing something again. I think my main idea is correct (the first two equations), but at the end I assume that in order to get this "normal" volume of the gas, it should be outside the tube, and its pressure should be equal to the outside pressure, so that's why I plug that into the formula.
I'd appreciate any hints you might throw at me.