2 mole of zinc is dissolved in HCl at 25- C- The work done in open vessel is-A- -2-447 kJB- -4-955 kJC- R - 2D- 3 R

The correct option is B 4.955 kJ The exn take place as follows 2Zn + 4HCl → 2ZnCl_2 + 2H_2 When H_2 gas is liberated in open vessel it pushes back its surrou ...

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Standard XII

Physics

Alpha Decay

2 mole of zin...

Question

2 mole of zinc is dissolved in HCl at $25^{\circ} C$ . The work done in open vessel is:

- 2.447 kJ

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- 4.955 kJ

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R/2

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Solution

The correct option is B
- 4.955 kJ

The exn take place as follows

$2 Z n + 4 H C l \to 2 Z n C l_{2} + 2 H_{2}$

When $H_{2}$ gas is liberated in open vessel it pushes back its surrounding and thus does p - V with in surroundings

$W = P_{e x t} Δ V = - P_{e x t} (V_{f} - V_{x})$

$V_{i} = 0$ thus $Δ V = V_{f}$

$V_{f} = \frac{n R T}{P_{e x t}}$

$W = - P e x t \times \frac{n R T}{P_{e x t}} = - n R T$

Given that $n = 2, R = 8.314 J / K m o l$

$T = 25 + 273 = 298 K$

$W = - 2 \times 8.314 \times 298 = 4955.144 J = 4.955 k J$

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