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 6635 J. What is the specific heat of the gas?

I believe this is done this way.

dE = q+w q is delta H and work is negative since the system did the work.
6635 = dH + (-346)
dH = 6635 + 346 = ?
dH = ? = mass x specific heat x (Tfinal-Tinitial)
Substitute and solve for specific heat.

To find the specific heat of the gas, we need to use the equation:

Q = mcΔT,

where Q is the heat energy transferred, m is the mass of the gas, c is the specific heat, and ΔT is the change in temperature.

In this case, we are given the following information:

m = 80.0 g (mass of the gas)
ΔT = 225 °C - 25 °C = 200 °C (change in temperature)
Q = 346 J (work done by the system) + 6635 J (increase in internal energy) = 6981 J (total heat energy transferred)

Now we can substitute these values into the equation and solve for c:

6981 J = (80.0 g)(c)(200 °C)

Dividing both sides by (80.0 g)(200 °C), we get:

c = 6981 J / (80.0 g)(200 °C)

Calculating this, we find:

c ≈ 0.436 J/g°C

Therefore, the specific heat of the gas is approximately 0.436 J/g°C.