Skip to content
##
vi. One mole of an ideal gas is compressed from 500 cm3 against a constant external pressure of 1.2 × 105 Pa. The work involved in the process is 36.0 J. Calculate the final volume. (200 cm3)

## Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution

### 4. Answer the following questions.

vi. One mole of an ideal gas is compressed from 500 cm3 against a constant external pressure of 1.2 × 105 Pa. The work involved in the process is 36.0 J. Calculate the final volume. (200 cm3)

Answer:-

- Home -
- vi. One mole of an ideal gas is compressed from 500 cm3 against a constant external pressure of 1.2 × 105 Pa. The work involved in the process is 36.0 J. Calculate the final volume. (200 cm3)

**Given:**

- Initial volume (\(V_1\)) = 500 cm
^{3} - External pressure (\(P_{\text{ext}}\)) = 1.2 × 10
^{5}Pa - Work (\(W\)) = 36.0 J

**To find:** Final volume (\(V_2\))

**Formula:** \(W = -P_{\text{ext}} \Delta V = -P_{\text{ext}} (V_2 - V_1)\)

**Calculation:**

- Initial volume (\(V_1\)) = 500 cm
^{3}= 0.5 dm^{3} - External pressure (\(P_{\text{ext}}\)) = 1.2 × 10
^{5}Pa = 1.2 bar - Work (\(W\)) = 36.0 J = 0.360 dm
^{3}bar (using \(1 \, \text{dm}^3 \, \text{bar} = 100 \, \text{J}\))

Now, from the formula:

\[W = -P_{\text{ext}} \Delta V = -P_{\text{ext}} (V_2 - V_1)\]Therefore:

\[\frac{0.360 \, \text{dm}^3 \, \text{bar}}{1.2 \, \text{bar}} = -(V_2 - 0.5 \, \text{dm}^3)\]So:

\[0.3 \, \text{dm}^3 = -V_2 + 0.5 \, \text{dm}^3\]Thus:

\[V_2 = 0.2 \, \text{dm}^3 = 200 \, \text{cm}^3\]The final volume (\(V_2\)) is 200 cm^{3}.

Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution