Name: Difference Between Ideal Gas and Real Gas
Uploaded: 2019-07-19
Duration: 6 min 16 s
Description: What is real gas? What is the difference between real gas and ideal gas? Watch this video to learn about this interesting concept of real gases.

Question

One mole of an ideal diatomic gas goes from $a$ to $c$ along the diagonal path as shown. The scale of the vertical axis is set by $p_{a b} = 5.0 kPa$ and $p_{c} = 0.50 kPa$ and the scale of the horizontal axis is set by $V_{b c} = 4.0 m^{3}$ and $V_{a} = 2.0 m^{3}$ . During the transition,

$(A)$ what is the change in the internal energy of the gas

$(B)$ how much energy is added to the gas as heat?

$(C)$ how much heat is required if the gas goes from $a$ to $c$ along the indirect path $a b c$ ?

Accepted Answer

For straight line on the $p V$ diagram, the work is given by area under the $p V$ graph,
$W_{straight} = (\frac{p_{i} + p_{f}}{2}) Δ V$ (for process $a \to c$ )
Now, internal energy is given by,
$E_{int} = n (\frac{f}{2}) R T = (\frac{f}{2}) p V$ , where $f$ is degree of freedom of diatomic gas.

The degrees of freedom for a diatomic gas is $(B)$
(a) The internal energy change, when gas goes from $a to c$ will be,
$∆ E_{int} = \frac{5}{2} (p_{c} V_{c} - p_{a} V_{a}) \Rightarrow ∆ E_{int} = \frac{5}{2} ((\frac{1}{2} \times 10^{3} Pa) (4.0 m^{3}) - (5.0 \times 10^{3} Pa) (2.0 m^{3}))$

$= - 20.0 \times 10^{3} J$

(b) The work done during the process represented by the diagonal path is

$W_{diag} = (\frac{p_{a} + p_{c}}{2}) (V_{c} - V_{a}) = (2.75 \times 10^{3} Pa) (2.0 m^{3})$

$\Rightarrow W_{diag} = 5.5 \times 10^{3} J .$

Consequently, the first law of thermodynamics gives
$Q_{diag} = Δ E_{int} + W_{diag} = (- 20.0 \times 10^{3} + 5.5 \times 10^{3}) J = - 14.5 \times 10^{3} J$

(c) The fact that $Δ E_{int}$ only depends on the initial and final states, and not on the details of the "path" between them, means we can write $Δ E_{a c} = Δ E_{a b c} = - 20.0 \times 10^{3} J$ for the indirect path, too. In this case, the work done consists of that done during the constant pressure part (the horizontal line in the graph) plus that done during the constant volume part (the vertical line):
$W_{indrect} = (5.0 \times 10^{3} Pa) (2.0 m^{3}) + 0 = 1.0 \times 10^{4} J$
Now, the first law of thermodynamics leads to
$Q_{indirect} = Δ E_{int} + W_{indirect} = (- 20.0 \times 10^{3} + 1.0 \times 10^{4}) J = - 10 \times 10^{3} J$

Important Questions on The Kinetic Theory of Gases

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