Question

3 moles of an ideal gas are expanded isothermally and reversibly from 10 m³ to 20 m³ at 300 K. Calculate the work done. (R = 8.314 KJ- mol-¹).​

Answer 1

Answer:

hshshshskeksjhehejekrkd

Answer 2

Final answer:

The work done = -5186.961 J

Given that: We are given 3 moles of an ideal gas, that are expanded isothermally and reversibly from 10 $m^{3}$ to 20 $m^{3}$ at 300 K.

R = 8.314 $J K^{-1} mol^{-1}$

To find: We have to find the work done.

Explanation:

• Suppose a system under consideration consists of an 'n' mole of an ideal gas.

Then the total work done (w) on the gas in an isothermal (T = constant) reversible expansion from the initial volume $V_{1}$ to $V_{2}$ of an ideal gas is:

$w_{rev} = -2.030\ nRT\ \log \frac{V_{2}}{V_{1}}$ ……….equation (I)

Where R is the universal gas constant.

• In an isothermal reversible expansion of an ideal gas, neither internal energy nor enthalpy changes (ΔU = 0 and ΔH = 0).

• The maximum work is considered as this isothermal reversible.

$w_{max}=w_{rev}$

Here:

R = 8.314 $J K^{-1} mol^{-1}$ [given]

$V_{1}$ = 10 $m^{3}$

$V_{2}$ = 20 $m^{3}$

T = 300 K

n = 3 mol

• Substitute these values in equation (I), we get the value of work done.

$w_{rev}=-2.030 \times 3\ mol \times 8.314 J K^{-1} mol^{-1} \times 300\ K \times \log (\frac{20\ m^{3}}{10\ m^{3}})$

$w_{rev} =(-2.030\times 3 \times 8.314 \times 300 \times \log 2) \ J$

• We know that $\log 2$ = 0.3010, substitute this value in $w_{rev}$.

$w_{rev} =(-2.030\times 3 \times 8.314 \times 300 \times 0.3010) \ J$

• Solving the above equation, we get:

$w_{rev}$ = -5186.961 J

Hence, work done = -5186.961 J

To know more about the concept please go through the links

https://brainly.in/question/13315514

https://brainly.in/question/14300031

#SPJ3