A gas is allowed to expand in a well insulated container against a constant external pressure of $2.5\, atm$ from an initial volume of $2.50\, L$ to a final volume of $4.50\, L$. The change in internal energy $\Delta U$ of the gas in joules will be

Question

Which is an adiabatic process?

Answer 1

We need to calculate the change in internal energy $\Delta U$ of a gas when it expands against a constant external pressure. The context involves concepts of thermodynamics, specifically the first law of thermodynamics for adiabatic processes.

Given:

Initial volume, $V_i = 2.50\, \text{L}$
Final volume, $V_f = 4.50\, \text{L}$
External pressure, $P_{\text{ext}} = 2.5\, \text{atm}$

Since the container is well insulated, it indicates that it is an adiabatic process, where no heat is exchanged with the surroundings ($q = 0$). According to the first law of thermodynamics:

\Delta U = q + w

Here, $q = 0$ (because it's an insulated system), so:

\Delta U = w

The work done by the gas when it expands is given by the formula:

w = -P_{\text{ext}} \times \Delta V

Where:

$P_{\text{ext}}$ is the external pressure
$\Delta V = V_f - V_i$ is the change in volume

Substitute the values:

$\Delta V = 4.50\, \text{L} - 2.50\, \text{L} = 2.00\, \text{L}$

Convert the volume from liters to cubic meters for SI units (since 1 L = 0.001 m³):

$\Delta V = 2.00 \times 0.001 = 0.002\, \text{m}^3$

For converting atm to Pascal (1 atm = 101325 Pa):

So, $P_{\text{ext}} = 2.5\, \text{atm} \times 101325\, \text{Pa/atm} = 253312.5\, \text{Pa}$

Substitute back into the work formula:

w = - (253312.5 \, \text{Pa}) \times (0.002 \, \text{m}^3)

w = -506.625\, \text{J}

Considering significant figures and rounding to the nearest joule, the change in internal energy $\Delta U$ is approximately:

\Delta U = -505\, \text{J}

Thus, the correct answer is -505 J.

Answer 2