In a constant-volume process, 211 J of energy is transferred by heat to 1.05 mol of an ideal monatomic gas initially at 303 K.
(a) Find the increase in internal energy of the gas.
J
(b) Find the work done on it.
J
(c) Find its final temperature.
K
i am having trouble with this question I tried using the formaula
change in eint=nCvchange in T. This is not working becasue i only have the initial temp. Please help explain?
To answer this question, you need to use the First Law of Thermodynamics, which states that the change in internal energy of a system is equal to the heat transferred to the system minus the work done by the system.
(a) To find the increase in internal energy of the gas, you can use the equation:
ΔEint = Q - W
Since the process is constant volume (V = constant), there is no work done (W = 0). Therefore, the equation simplifies to:
ΔEint = Q
Thus, the increase in internal energy of the gas (ΔEint) equals the heat transferred (Q).
ΔEint = 211 J
(b) Now let's find the work done on the gas. In a constant-volume process, the volume remains constant, so no work is done.
W = 0 J
(c) Finally, let's determine the final temperature of the gas. You mentioned that you only have the initial temperature, but in the constant-volume process, the temperature of an ideal gas is directly proportional to its internal energy. Therefore, the equation for the change in internal energy can be rewritten as:
ΔEint = nCvΔT
Where:
- n is the number of moles of the gas (given as 1.05 mol)
- Cv is the molar specific heat at constant volume for an ideal monatomic gas (given as 3/2R)
- ΔT is the change in temperature
Now, we can rearrange the equation to solve for ΔT:
ΔT = ΔEint / (nCv)
ΔT = 211 J / (1.05 mol * (3/2 R))
To proceed, it is important to know the value of the gas constant (R), which depends on the unit of temperature. If you let me know the unit, I can help you calculate the final temperature (Tf).