Question

Find the heat added to or taken from the gas in part (a) if it

expands at a pressure of 110 kPa from a volume of 0.77 m^3 to a
volume of 0.96 m^3

Answer 1

The change in internal energy
`(\( \Delta U \))` of the gas during the process is
`\(-20900 \, \text{J}\).`

To find the heat added to or taken from the gas during its expansion, we can use the first law of thermodynamics, which states:

`\[ \Delta U = Q - W \]`

Where:

-
`\( \Delta U \)` is the change in internal energy of the gas.

-
`\( Q \)` is the heat added to or taken from the gas.

-
`\( W \)` is the work done by the gas.

Given that the gas expands at a constant pressure of 110 kPa from an initial volume of 0.77 m³ to a final volume of 0.96 m³, the work done by the gas during expansion
`(\( W \))` can be calculated using the formula:

`\[ W = P \cdot \Delta V \]`

Where:

- P is the pressure (110 kPa).

-
`\( \Delta V \) is the change in volume (\( V_{\text{final}} - V_{\text{initial}} \)).`

Given:

-
`\( P = 110 \)` kPa

-
`\( V_{\text{initial}} = 0.77 \)` m³

-
`\( V_{\text{final}} = 0.96 \)` m³

`\[ \Delta V = V_{\text{final}} - V_{\text{initial}} \]`
`\[ = 0.96 \, \text{m³} - 0.77 \]`m³

Now, let's calculate the work done by the gas:

`\[ W = P \cdot \Delta V = 110 \, \text{kPa} * 0.19 \, \text{m³} \]`

To perform the calculation, let's convert kPa to Pa (1 kPa = 1000 Pa):

`\[ W = 110 * 10^3 \, \text{Pa} * 0.19 \, \text{m³} = 20900 \, \text{J} \]`

The work done by the gas during expansion is
`\( 20900 \, \text{J} \).`

Now, using the first law of thermodynamics:

`\[ \Delta U = Q - W \]`

Since the process is adiabatic (no heat exchange), Q = 0. Therefore:

`\[ \Delta U = -W = -20900 \, \text{J} \]`