Answer:
ΔU = Q - W = 108
Step-by-step explanation:
We can solve this problem using the First Law of Thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:
ΔU = Q - W
where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.
To solve for ΔU, we need to first calculate the heat added to the system. We can use the formula for electrical power to do this:
P = IV
where P is the power (in watts), I is the current (in amperes), and V is the voltage (in volts).
We are given that the electric heater operates at 150 W for 12.0 min, or 720 s. Assuming a voltage of 120 V (typical for a household circuit), we can calculate the current as:
I = P/V = 150 W / 120 V = 1.25 A
Using Ohm's Law (V = IR), we can also calculate the voltage drop across the heater as:
V = IR = 1.25 A * R
where R is the resistance of the heater. We are not given the resistance directly, but we can calculate it using the formula for electrical power:
P = V^2/R
R = V^2/P = (1.25 A * R)^2 / 150 W
Solving for R gives:
R = 1.5625 Ω
Now that we know the current and resistance of the heater, we can calculate the heat added to the gas as:
Q = Pt = VI*t = (1.25 A * 120 V) * (720 s) = 108000 J
where t is the time (in seconds).
Next, we need to calculate the work done by the gas. Since the gas is expanding against a constant pressure, the work done is:
W = -PΔV
where P is the pressure (in atm) and ΔV is the change in volume (in liters). We are given that the pressure is constant at 1.03 atm, and the volume changes from 3.00 L to 11.0 L, so:
ΔV = 11.0 L - 3.00 L = 8.00 L
Converting liters to cubic meters (m^3) and using the ideal gas law (PV = nRT), we can calculate the initial and final number of moles of gas:
n1 = (P1V1) / (RT1) = (1.03 atm * 3.00 L) / (0.08206 Latm/molK * 300 K) = 0.1266 mol
n2 = (P2V2) / (RT2) = (1.03 atm * 11.0 L) / (0.08206 Latm/molK * 300 K) = 0.4649 mol
where T is the temperature (in Kelvin), which we assume remains constant.
Using the equation for the work done by an ideal gas in a reversible process:
W = -nRT ln(V2/V1)
where ln is the natural logarithm, we can calculate the work done by the gas as:
W = -nRT ln(V2/V1) = -(0.1266 mol) * (0.08206 Latm/molK) * 300 K * ln(11.0 L / 3.00 L) = -951 J
Finally, we can calculate the change in internal energy of the gas as:
ΔU = Q - W = 108