Maqboole Posted March 1, 2008 Share Posted March 1, 2008 Hi there I hope you people must be doing well & if you be kind enough to please help me in my problem I would be grateful thx. Maqbool This is Physics question related to ideal gas Q: A uniform capillary tube, closed at one end, contained air trapped by tread of mercury 85mm long. when tube was held horizontally , the length of the air column was 50mm. when it was held vertically with a corked end downward The length was 45mm. find the atmospheric pressure. take g=10 Density of mercury; 14 x 10^3 How to solve it I tried it answer comes 5, use p=density x gravity x change in length Ans should be 1.01 x 10^ 5 (according to Mark scheme). PS. I need urgent help please, Ok don't give the answer but at least direct me to the solution! Thanks! Link to comment Share on other sites More sharing options...
swansont Posted March 1, 2008 Share Posted March 1, 2008 I'm not sure I understand the scenario. It looks like you have a column of mercury, with trapped air on the closed side of the tube, and trapped air below because of the cork. The air below the mercury must have sufficient pressure to hold up the mercury and the pressure above. The air pockets started at atmosphere. But you give the length as 50 mm when the pressure is at atmosphere, and 45 mm when the tube is inverted ("corked end downward"). The air column should be longer, not shorter. So either I'm misunderstanding something, or the question is worded incorrectly. Link to comment Share on other sites More sharing options...
Maqboole Posted March 1, 2008 Author Share Posted March 1, 2008 This is the same problem I have actually. I think I need to relook the question and tell whether I have stated wrongly or not. My apologies if I have though! Link to comment Share on other sites More sharing options...
iNow Posted March 1, 2008 Share Posted March 1, 2008 I think I need to relook the question and tell whether I have stated wrongly or not. Come back and let us know when you do, okay? I'd like to see how this question resolves. Link to comment Share on other sites More sharing options...
John Cuthber Posted March 1, 2008 Share Posted March 1, 2008 The question looks OK to me. With the tube horizontal the air column is at one atmosphere and is 50mm long. With the sealed end downwards the mercury squashes the air to 45mm long so the pressure must have risen to 50/45 atmospheres when you added the pressure due to the mercury. Link to comment Share on other sites More sharing options...
swansont Posted March 1, 2008 Share Posted March 1, 2008 The question looks OK to me. With the tube horizontal the air column is at one atmosphere and is 50mm long. With the sealed end downwards the mercury squashes the air to 45mm long so the pressure must have risen to 50/45 atmospheres when you added the pressure due to the mercury. OK, then my mental image of the capillary tube was wrong. I was thinking something open only at one end (and being corked, not open at either end) Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now