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 solve this problem, you can use the first law of thermodynamics, which states that the change in internal energy (ΔEint) of a system is equal to the heat transferred (Q) into the system minus the work (W) done by the system:
ΔEint = Q - W
(a) To find the increase in internal energy of the gas, we can use the formula:
ΔEint = Q
Given that the heat transferred (Q) is 211 J, the increase in internal energy of the gas is also 211 J.
(b) To find the work done on the gas, we need to use the equation:
W = -PΔV
However, since this is a constant-volume (isochoric) process, the volume remains constant, and as a result, no work is done on the gas. Therefore, the work done on the gas in this case is 0 J.
(c) To find the final temperature of the gas, we can rearrange the ideal gas equation:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
In a constant-volume process, the volume remains constant, and therefore the pressure does not change. Therefore, we can rewrite the equation as:
P1V1 = P2V2
Since V1 = V2, we can simplify the equation as:
P1 = P2
Given that this is an ideal monatomic gas, we can use the ideal gas law equation for an ideal monatomic gas:
PV = nRT
We can rearrange this equation to solve for the final temperature (T2) as follows:
T2 = (P2V2) / (nR)
However, we still need the value of the pressure (P2) to find the final temperature.
Unfortunately, the information provided does not give us enough information to directly calculate the final temperature, as we need either the final pressure or the final volume. Without this information, we cannot determine the final temperature of the gas.