C. Thermodynamic processes 5. A thermodynamic system gains \( 2,700 \mathrm{~J} \) of heat at a constant pressure of 1.3 x \( 10^{5} \mathrm{~Pa} \) and its internal energy increases by 3990 J . What is the change in volume of the system, and is it an increase or a decrease?

To find the change in volume (\( \Delta V \)) of the thermodynamic system, we can use the first law of thermodynamics, which states: \[ \Delta U = Q - W \] Where: - \( \Delta U \) = change in internal energy (3990 J) - \( Q \) = heat added to the system (2700 J) - \( W = P \Delta V \) = work done by the system First, let's rearrange the equation to solve for work: \[ W = Q - \Delta U \] Substituting the known values: \[ W = 2700 \, \text{J} - 3990 \, \text{J} = -1290 \, \text{J} \] Since work is done on the system, we have: \[ - P \Delta V = -1290 \, \text{J} \] Now, substituting in the pressure (\( P = 1.3 \times 10^5 \, \text{Pa} \)): \[ 1.3 \times 10^5 \Delta V = 1290 \] Now, solving for \( \Delta V \): \[ \Delta V = \frac{1290}{1.3 \times 10^5} \approx 0.00992 \, \text{m}^3 = 9.92 \, \text{L} \] Since \( \Delta V \) is positive, it indicates an increase in volume. So, the change in volume is approximately \( 9.92 \, \text{L} \) and it is an increase!