A fixed mass of an ideal gas slowly releases 1500 j of heat and as a result contract slowly,at a constant pressure 2.0 x 104 pa, from a volume of 0.025 M3. What is the effect on the internal energy of the gas?
A it decrease by 2000 j.
B it decreases by 1000 j.
C it is unchanged.
D it increase by 1000 j
To determine the effect on the internal energy of the gas, we can make use of the First Law of Thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.
The equation can be written as:
ΔU = Q - W
Where:
ΔU is the change in internal energy,
Q is the heat added to the system,
W is the work done by the system.
In this case, the system is experiencing a constant pressure process, which means the work done by the system can be calculated using the equation:
W = P * ΔV
Where:
P is the constant pressure,
ΔV is the change in volume.
Given the following information:
Q = 1500 J (heat added to the system),
P = 2.0 x 10^4 Pa (constant pressure),
ΔV = (final volume - initial volume) = 0.025 m^3 (final volume) - 0.025 m^3 (initial volume) = 0 (since the volume remains constant).
Plugging the values into the equations, we can calculate the work done by the system:
W = P * ΔV = (2.0 x 10^4 Pa) * 0 = 0 J
Now, we can determine the change in internal energy:
ΔU = Q - W = 1500 J - 0 J = 1500 J
Therefore, the change in internal energy of the gas is 1500 J.
However, none of the provided answer choices match this value. The correct answer should be "None of the above" or "Not enough information provided."