when 1240 J of heat are added to one mole of an ideal monatomic gas its temperature increases from 273K to 277K. Find the work done by the gas during the process.
To find the work done by the gas during this process, we can use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat (Q) added to the system minus the work (W) done by the system:
ΔU = Q - W
In this case, we are given the heat added (Q) as 1240 J and the temperature change (ΔT) as 277K - 273K = 4K.
Now, for an ideal monatomic gas, the change in internal energy (ΔU) is given by the equation:
ΔU = 3/2 * n * R * ΔT
where n is the number of moles of gas and R is the ideal gas constant.
Since we are given that there is one mole of gas, n = 1 mol, and for an ideal monatomic gas, R = 8.314 J/(mol·K).
Thus, we can calculate the change in internal energy (ΔU):
ΔU = 3/2 * 1 mol * 8.314 J/(mol·K) * 4K = 49.884 J
Now, we substitute the known values into the first law of thermodynamics equation:
49.884 J = 1240 J - W
Solving for W, the work done by the gas:
W = 1240 J - 49.884 J = 1190.116 J
Therefore, the work done by the gas during the process is approximately 1190.116 J.