Work done (in bar-L) on the gas in reversible isothermal compression is : | TestHub

Passage / Comprehension

If the boundary of a system moves by an infinitesimal amount, the work involved is given by:

$d w = - P_{e x t} d V$

For an irreversible process:

$w = - P_{e x t} Δ V$ (where $Δ V = V_{f} - V_{i}$ )

For a reversible process:

$P_{e x t} = P_{in t} \pm d P \approx P_{in t}$

So, for a reversible isothermal process:

$w = - n RT ln \frac{V _{f}}{V _{i}}$

Two moles of an ideal gas undergo isothermal compression along three different paths:

(i) Reversible compression from $P_{i} = 2 ba r$ and $V_{i} = 8 L$ to $P_{f} = 20 ba r$ .

(ii) A single-stage compression against a constant external pressure of $20 ba r$ .

(iii) A two-stage compression consisting initially of compression against a constant external pressure of $10 ba r$ until $P_{g a s} = P_{e x t}$ , followed by compression against a constant pressure of $20 ba r$ until $P_{g a s} = P_{e x t}$ .

Work done (in bar-L) on the gas in reversible isothermal compression is :

Options:

Answer:

Solution:

For reversible isothermal compression:

$w = - n RT ln (\frac{V _{f}}{V _{i}})$

Using the isothermal condition:

$P_{i} V_{i} = P_{f} V_{f}$

$V_{f} = \frac{P _{i} V _{i}}{P _{f}} = \frac{2 \times 8}{20} = 0.8 L$

Now,

$w = - P_{i} V_{i} ln (\frac{V _{f}}{V _{i}})$

$w = - 16 ln (\frac{0.8}{8}) = - 16 ln (0.1) = 16 ln (10)$

$w = 16 \times 2.303 = 36.848 ba r - L$

Hence, the work done is $36.848 ba r - L$ .

Chemistry - Thermodynamics - I Question with Solution | TestHub

Options:

Doubts & Discussion