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`        Please help me to solve this question. It's urgent.`
one year ago

```							Initial Mass, W1 = 13gFinal Mass = W2Initial Temperature, T1 = 27℃ = 300KFinal Temperature, T2 = 52℃ = 325KIt is given that initial and final pressure are same.So, Initial Pressure = Final Pressure = P PV = nRT In initial casePV = nRT  INITIAL CASEPV = nRT1, ​​​ and n = weight/molecular mass= W1/MSo, PV = (W1RT1)/MTherefore, MV = (W1RT1)/P, substituting the valuesMV = (13 × R × 300)/PMV = 3900R/P ------------> Eq.1 FINAL CASEPV = nRT, and n=W2/MPV = (W2RT2)/MPVM = W2RT2Substituting the value of MV from Equation 1, and value of T2 we get,P × (3900R)/P = W2 × R × 325W2 = 3900/325W2 = 12gWeight to be removed = W1 - W2= 13 - 12 = 1g.Hope you understood it.
```
one year 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 TT1 = 300 K ; T2=325 K d 1 T 1= d 2 T2d2 = d1T1/T2 = 300/325 d1=12/13 d1 Initial Volume  V= 13/d1Final 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 Ask ittians expert B Tech Chem
```
one year ago
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