ThermodynamicsHard

Question

Two rigid adiabatic vessels A (volume = 4 L) and B (volume = 6 L), which initially contains two gases at different temperatures are connected by a pipe of negligible volume. The vessel A contains 2 moles of Ne gas (Cp,m = 5 cal/K-mol) at 300 K and vessel B contains 3 moles of SO2 gas (Cp,m = 8 cal/K-mol) at 400 K. What is the final pressure (in atm) when the valve is opened and 12 kcal heat is supplied through it to the vessels? (R = 0.08 L-atm/K-mol)

Options

A.3.5
B.7
C.35
D.70

Solution

$q = \left( n.C_{V,m}.\Delta T \right)_{Ne} + \left( n.C_{V,m}.\Delta T \right)_{SO_{3}}$

$\text{Or, }12 \times 10^{3} = 2 \times 3 \times \left( T_{f} - 300 \right) + 3 \times 6 \times \left( T_{f} - 400 \right) \Rightarrow T_{f} = 875\text{ K} $$$\text{Now, }P_{\text{total}} = \frac{nRT}{V} = \frac{5 \times 0.08 \times 875}{10} = 35\text{ atm}$$

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