Two glass bulbs A (of 100 mL capacity), and B (of 150 mL capacity) containing same gas are connected by
a small tube of negligible volume. At particular temperature the pressure in A was found to be 20 times more
than that in bulb B. The stopcock is opened without changing the temperature. The pressure in A will
(A) drop by 75% (B) drop 57% (C) drop by 25% (D) will remain same
Answers
Answer:
57% drop.
Explanation:
If we equate the number moles in both the system so we will get that:-
Let n= number of moles.
n(in A) + n(is B) =n(in A) + n(is B).
Where left hand is initial moles and the right hand is final mole. We also know that PV=nRT.
So, P(A)V(A)/RT + P(B)V(B)/RT = P(A)V(1)/RT + P(A)V(2)/RT.
So, 100P(A) + 150P(B)/RT = P(A)V1 + P(A)V2/RT.
Since we know that the volume P(A) becomes equal to P(B) when the volumes are in V1 and V2.
Now, 50(2P(A) + 3P(B)) = P(A)(V1+V2).
So, on solving we will get the value of P(A) = 43/5P(B).
Hence, we know that P(A) = 20P(B) which is the initial pressure.
Also, P(A)=43/5P(B) which is the final pressure.
So, the change in pressure = 20-43/5 = 11.4 atmospheric pressure,
Therefore, the percentage drop = 100*11.4/20 = 57% of drop.
Answer:
Explanation:
Let the original pressure in B be P.
∴ Pressure in A = 20 P.
Given that :
Volume in A = 100 mL
Volume in B = 150 mL
After opening the stopcock,
total volume =(100+150)mL=250mL
According to Boyle's law :
P
f
V
f
=P
i
V
i
Let the partial pressure of gas in A be P
A
:
P
A
×250=100×20P
⇒P
A
=
250
100×20P
=8P
Let the Partial pressure of gas in B be P
B
:
P
B
×250=150×P
P
B
=
250
150P
=0.6P
∴ Total pressure =8P+0.6P=8.6P