Question

A sample of gas at room temperature is placed in an evacuated bulb of volume 0.51 dr and is found to
exert a pressure of 0.24 atm. The bulb is connected to another evacuated bulb whose volume is 0.63 dm.
The new pressure of the gas at room temperature will be:
(A) 1.073 atm
(B) 1.1073 atm
(C) 2.1073 atm
(D) 0.1073 atm
​

Answer 1

Answer: (D) 0.1073 atm

​Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$     (At constant temperature and number of moles)

$P_1V_1=P_2V_2$  

where,

$P_1$ = initial pressure of gas  = 0.24 atm

$P_2$ = final pressure of gas  = ?

$V_1$ = initial volume of gas  = $0.51dm^3$

$V_2$ = final volume of gas  = $(0.51+0.63)dm^3=1.14dm^3$

$0.24\times 0.51=P_2\times 1.14$  

$P_2=0.1073atm$

Therefore, the new pressure of the gas at room temperature will be 0.1073 atm

Answer 2

Explanation:

p1=0.24 atm

V1=0.51 dm3

V2=0.51+0.63=1.14dm3

p2=?

p1v1=p2v2

0.24*0.51=p2*1.14

p2=0.24*0.51/1.14

p2=0.1224/1.14

p2=0.1073atm

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