Question

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 △U of the gas in joules will be

• A. 1136.25 J
• B. - 500 J
• C. –505 J
• D. +505 J

Answer 1

The Correct Option is C

To find the change in internal energy (\(\Delta U\)) of the gas, we use the concept of work done by the gas and the fact that the process is adiabatic (since the container is well insulated) meaning the heat exchange (\(q\)) is zero.

The formula for the work done by the gas during expansion against constant external pressure is given by: 

\(W = - P_{\text{ext}} \times \Delta V\)

Where:

• \(P_{\text{ext}} = 2.5 \, \text{atm}\) is the external pressure.
• \(\Delta V = V_{\text{final}} - V_{\text{initial}} = 4.50 \, \text{L} - 2.50 \, \text{L} = 2.00 \, \text{L}\)

Substituting the values, we have:

\(W = - 2.5 \, \text{atm} \times 2.00 \, \text{L}\)

Converting atm-L to Joules (1 atm-L = 101.3 J):

\(W = - 2.5 \times 2.00 \times 101.3 = - 505 \, \text{J}\)

According to the first law of thermodynamics, the change in internal energy (\(\Delta U\)) is given by:

\(\Delta U = q + W\)

For an adiabatic process, \(q = 0\), hence:

\(\Delta U = 0 + (-505) = -505 \, \text{J}\)

Thus, the change in internal energy of the gas is -505 J.

Therefore, the correct answer is -505 J, which corresponds to the option:

–505 J

.