Three rigid adiabatic vessels $A, B & C$ which initially contain three different gases at different temperatures are connected by connecting tube of negligible volume. The vessel $A$ contain $1$ mole Helium gas, at $300 K$ , vessel $B$ contain $2$ mole $S O_{2}$
gas at $400 K$ and vessel $C$ contain $3$ mole $C O_{2}$ gas at temperature $500 K$ . What is the final pressure (in atm) attained by gases when tube connecting three vessels is opened and additional $16.8$ kcal heat supplied to vessel. The volume of $A, B & C$ vessel is $1, 2 & 3$ litre respectively.
Use: $R = 2 c a l o r i e / m o l - K$ ( for temperature calculation)
$R = 0.08 l i t r e - a t m / m o l e - K$ ( for pressure calculation)
$C_{v} (H e) = \frac{3}{2} R, C_{v} (S O_{2}) = 3 R & C_{v} (C O_{2}) = \frac{5}{2} R$

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$4 \times 10^{- 2}$

$8 \times 10^{- 1}$

$4 \times 10^{2}$

$8.0 \times 10^{1}$

answer is D.

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Detailed Solution

$C_{v} (H e) = \frac{3}{2} R, C_{v} (S O_{2}) = 3 R & C_{v} (C O_{2}) = \frac{5}{2} R$

Therefore $\frac{3}{2} R (T_{f} - 300) + 2 \times 3 R (T_{f} - 400) + 3 \times \frac{5}{2} R (T_{f} - 500) = 16800$

$T_{f} \{\frac{3}{2} R + 6 R + \frac{15}{2} R\} = 16800 + R \{\frac{3}{2} (300) + 6 \times 400 + 3 \times \frac{5}{2} \times 500\} T_{f} R [\frac{30}{2}] = 16800 + R \{\frac{900 + 7500 + 4800}{2}\} T_{f} = 1000 K P_{f} = \frac{6 \times 0.8 \times 1000}{6} = 80 a t m = 8.0 \times 10^{1}$