An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its internal energy increased by 8305 J. What is the specific heat of the gas?
dE = q + w
You know dE and work. Solve for q.
Then q = mass gas x specific heat gas x (Tfinal-Tinitial)
Substitute and solve for speific heat.
To determine the specific heat capacity of a gas, we can use the equation:
ΔQ = m * c * ΔT
Where:
ΔQ is the change in heat energy,
m is the mass of the gas,
c is the specific heat capacity, and
ΔT is the change in temperature.
In this case, we are given:
m = 80.0 grams,
ΔT = (225 °C - 25 °C) = 200 °C,
ΔQ = 346 J, and
ΔU = 8305 J.
We can use the relationship between heat and change in internal energy:
ΔQ = ΔU + W
Where:
ΔU is the change in internal energy, and
W is the work done by the system.
Substituting the given values into this equation, we have:
346 J = 8305 J + W
Rearranging the equation, we find:
W = 346 J - 8305 J = -7959 J
Given that the work done by the system is negative, it means work is done on the system.
Now, let's find the value of c. Rearranging the equation ΔQ = m * c * ΔT, we have:
c = ΔQ / (m * ΔT)
Substituting the known values:
c = (-7959 J) / (80.0 g * 200 °C)
c = -0.49875 J/(g * °C)
Therefore, the specific heat capacity of the gas is approximately -0.49875 J/(g * °C).