A 7.56-g sample of gas is in a balloon that has a volume of 10.5 L. Under an external pressure of 1.05 atm, the balloon expands to a volume of 15.00 L. Then the gas is heated from 0.00 degree C to 25 C. If the specific heat of the gas is 0.909 J/g*C, what are work, heat, and change E for the overall process?
To find the work, heat, and change in internal energy (ΔE) for the overall process, we need to calculate each one separately.
1. Work (W):
The work done by the gas can be calculated using the formula:
W = -PΔV
where W is the work, P is the pressure, and ΔV is the change in volume.
In this case, the initial volume (V1) is 10.5 L, and the final volume (V2) is 15.00 L. The pressure (P) is given as 1.05 atm.
So, ΔV = V2 - V1 = 15.00 L - 10.5 L = 4.5 L.
Substituting the values:
W = -PΔV = -(1.05 atm)(4.5 L) = -4.725 atm·L.
2. Heat (Q):
The heat absorbed or released by the gas can be calculated using the formula:
Q = mcΔT
where Q is the heat, 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 specific heat (c) as 0.909 J/g·°C and the change in temperature (ΔT) as 25 °C.
To calculate the mass (m) of the gas, we can use the given sample's mass of 7.56 g.
Substituting the values:
Q = (7.56 g)(0.909 J/g·°C)(25 °C) = 173.4445 J.
3. Change in internal energy (ΔE):
The change in internal energy can be calculated using the formula:
ΔE = Q - W
where ΔE is the change in internal energy, Q is the heat, and W is the work.
Substituting the calculated values:
ΔE = 173.4445 J - (-4.725 atm·L) = 178.1695 J.
Therefore, the work (W) is -4.725 atm·L, the heat (Q) is 173.4445 J, and the change in internal energy (ΔE) is 178.1695 J for the overall process.