logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

A jar contains a gas and a few drops of water at 'T' K . The pressure in the jar is 830mm of Hg. The temperature of the jar is reduced by $1\%$. The vapour pressure of water at two temperature are 30 and 25 mm of Hg. Calculate the new pressure in the jar.

$(a)\;817mm\;of\;Hg\qquad(b)\;8.17mm\;of\;Hg\qquad(c)\;817cm\;of\;Hg\qquad(d)\;8.17 \;litre$

Can you answer this question?
 
 

1 Answer

0 votes
At 'T' K
$P_{gas} = P_{dry gas} + P_{moisture}$
$P_{dry gas} = 830 - 30$
$\;\;\;\;\;\;\;\;\;\;\;= 800 mm$
Now at new temperature $T_1 = T - \large\frac{T}{100}$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 0.99 T$
Since $V_1 = V_2 \large\frac{P}{T}$ = constant
$P_{dry gas} = \large\frac{800\times0.99T}{T}$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\; =792 mm$
$P_{gas} = P_{dry gas}+P_{moisture}$
$\;\;\;\;\;\;\;\;\;=792 + 25$
$\;\;\;\;\;\;\;\;\;\;\;= 817 \;mm\;of\;Hg$
Hence answer is (A)
answered Apr 1, 2014 by sharmaaparna1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...