One mole of an ideal gas at 900 K- undergoes two reversible processes- I followed by II- as shown below-If the work done by the gas in the two process are same- the value of lnV3V2U-internal energy- S-entropy- p-pressure- V-volume- R-gas constant Given- molar heat capacity at constant volume- Cv-m of the gas is 52R -

1^st nbsp;process is adiabatic since entropy is constant. W_1=Δ U Δ U=450R 2250R= 1800R W_1= 1800R nbsp; nbsp; ...(1) In 2^nd process internal energy is ...

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Byju's Answer

Standard XII

Chemistry

Adsorption and Absorption

One mole of a...

Question

One mole of an ideal gas at $900 K$ , undergoes two reversible processes, $I$ followed by $I I$ , as shown below.
If the work done by the gas in the two process are same, the value of $ln \frac{V_{3}}{V_{2}}$

$(U : internal energy, S : entropy, p : pressure, V : volume, R : gas constant)$
( Given: molar heat capacity at constant volume, $C_{v, m}$ of the gas is $\frac{5}{2} R$ )
.

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Solution

$1^{s t}$ process is adiabatic since entropy is constant.
$W_{1} = Δ U$
$Δ U = 450 R - 2250 R = - 1800 R$
$W_{1} = - 1800 R$ ...(1)

In $2^{n d}$ process internal energy is constant it means it is a isothermal process.
$W_{2} = - 2303 n R T log \frac{V_{3}}{V_{2}}$ ...(2)
$= - n R T l n \frac{V_{3}}{V_{2}}$ ...(3)
Given, $n = 1$ mole, here temperature is unknown
$U = n C_{v} T$ for process $I I$
$450 R = 1 \times \frac{5}{2} R T$
$T = \frac{450 \times 2}{5} = 180 K$
Equation (1) = Equation (2)
$W_{1} = W_{2}$
$- 1800 R = - 1 \times R \times 180 l n \frac{V_{3}}{V_{2}}$
$l n \frac{V_{3}}{V_{2}} = \frac{1800}{180} = 10$

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