Please help me to solve this question. It's urgent.
Sahani Kumar , 6 Years ago
Grade 12th pass
Mechanics> Please help me to solve this question. It...
2 Answers
Aniruddha Dewangan
Last Activity: 6 Years ago
Initial Mass, W1 = 13g
Final Mass = W2
Initial Temperature, T1 = 27℃ = 300K
Final Temperature, T2 = 52℃ = 325K
It is given that initial and final pressure are same.
So, Initial Pressure = Final Pressure = P
PV = nRT
In initial case
PV = nRT
INITIAL CASE
PV = nRT1, and n = weight/molecular mass= W1/M
So, PV = (W1RT1)/M
Therefore, MV = (W1RT1)/P, substituting the values
MV = (13 × R × 300)/P
MV = 3900R/P ------------> Eq.1
FINAL CASE
PV = nRT, and n=W2/M
PV = (W2RT2)/M
PVM = W2RT2
Substituting the value of MV from Equation 1, and value of T2 we get,
P × (3900R)/P = W2 × R × 325
W2 = 3900/325
W2 = 12g
Weight to be removed = W1 - W2
= 13 - 12 = 1g.
Hope you understood it.
Atul Nagar
Last Activity: 6 Years ago
Key Idea: When the gas is heated its density reduces. Due to this volume increases. To keep the same volume and pressure some of the filled gas need to be relased to compensate for density increase.
Solution : density d is inversely proportional to T
T1 = 300 K ; T2=325 K
d 1 T 1= d 2 T2
d2 = d1T1/T2 = 300/325 d1=12/13 d1
Initial Volume V= 13/d1
Final Volume = m / d2 = m/ (12/13d1)
equating both volume we have , m = 12 g
Mass need to be reduced = 13 g – 12 g = 1 g
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