A bulb of unknown volume V contains a gas at 1 atm pressure. This bulb...
**Solution:**
To solve this problem, we can use the ideal gas law equation:
PV = nRT
where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Since the temperature remains constant, we can simplify the equation to:
PV = constant
Let's analyze the situation step by step:
1. Initial conditions: The first bulb contains a gas at 1 atm pressure. The volume of this bulb (V1) is unknown.
2. Connection: The first bulb is connected to another evacuated bulb of volume 0.5 L.
3. Opening the stopcock: When the stopcock is opened, the pressure in each bulb becomes 7.58 * 10^4 Pa. Let's call this common pressure Pfinal.
Now, let's apply the ideal gas law to both bulbs before and after the stopcock is opened:
**First Bulb (Initial Conditions):**
P1 * V1 = n * R * T
**First Bulb (After Opening the Stopcock):**
Pfinal * V1 = n * R * T
**Second Bulb (After Opening the Stopcock):**
Pfinal * V2 = n * R * T
Since the number of moles and the temperature remain constant, we can equate the equations for the first and second bulbs:
Pfinal * V1 = Pfinal * V2
Now, let's substitute the values:
Pfinal = 7.58 * 10^4 Pa
V2 = 0.5 L
7.58 * 10^4 * V1 = 7.58 * 10^4 * 0.5
V1 = 0.5 L
Therefore, the volume of the first bulb (V) is 0.5 L.
In conclusion, the unknown volume of the first bulb is 0.5 L.
A bulb of unknown volume V contains a gas at 1 atm pressure. This bulb...
Boyle's law.... P1v1=p2v2
Here p1=1atm=1.013*10^5paV1=v, p2=7.58*10^5 paAnd v2=(v+0.5) litres. Substituting the value of p1, v1, p2 and v2 in the above relation. 1.013*10^5*v=7.58*10^5*(0.5+v)or
v=- 0.586 litres
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