An ideal gas expands against a constant external pressure of 2 atmosphere from 20 litre to 40 litre and absorbs 10KJ of heat from surrounding. What is the change in internal energy of the system?A. 4052JB. 5948JC. 14052JD. 9940J

To calculate the change in internal energy (ΔU\Delta U) of the system, we need to use the first law of thermodynamics, which is given by:
ΔU=Q−W\Delta U = Q - W
Where:
• ΔU\Delta U is the change in internal energy,
• QQ is the heat absorbed by the system,
• WW is the work done by the system.
Step 1: Calculate the work done by the gas
The work done by an ideal gas expanding against a constant external pressure is given by:
W=PextΔVW = P_{\text{ext}} \Delta V
Where:
• Pext=2 atmP_{\text{ext}} = 2 \, \text{atm} (external pressure),
• ΔV=Vfinal−Vinitial=40 L−20 L=20 L\Delta V = V_{\text{final}} - V_{\text{initial}} = 40 \, \text{L} - 20 \, \text{L} = 20 \, \text{L}.
Now, we need to convert the pressure from atmospheres to pascals and the volume from liters to cubic meters for consistency with SI units:
• 1 atm=101325 Pa1 \, \text{atm} = 101325 \, \text{Pa},
• 1 L=1×10−3 m31 \, \text{L} = 1 \times 10^{-3} \, \text{m}^3.
So,
W=2 atm×20 L=2×101325 Pa×20×10−3 m3=4053 J.W = 2 \, \text{atm} \times 20 \, \text{L} = 2 \times 101325 \, \text{Pa} \times 20 \times 10^{-3} \, \text{m}^3 = 4053 \, \text{J}.
Step 2: Use the first law of thermodynamics
The heat absorbed by the system is given as Q=10 kJ=10000 JQ = 10 \, \text{kJ} = 10000 \, \text{J}.
Now, using the first law of thermodynamics:
ΔU=Q−W=10000 J−4053 J=5947 J.\Delta U = Q - W = 10000 \, \text{J} - 4053 \, \text{J} = 5947 \, \text{J}.
Final Answer:
The change in internal energy of the system is approximately 5948 J.
So, the correct answer is B. 5948 J.