Let 70.0 J of heat be added to a particular ideal gas. As a result, its volume changes from 65.0 cm3 to 169.0 cm3 while the pressure remains constant at 1.7 atm. By how much did the internal energy of the gas change?
P*deltaV= external work -work internal
1.7atm*104cm^3=-work done internally+70J
101.3E3*1.7*104E-6 J-70J=-work done internally
work internal=-.1013*1.7*104*6+70J
= -17.9+70=52.1 J
check that.
To calculate the change in internal energy of the gas, we can use the formula:
ΔU = Q - W
Where:
ΔU = change in internal energy of the gas
Q = heat added to the gas
W = work done by the gas
In this case, the pressure remains constant, so the work done (W) is zero. Therefore, the change in internal energy (ΔU) is equal to the heat added (Q).
Given that Q = 70.0 J, the change in internal energy of the gas is also 70.0 J.
To determine the change in internal energy of the gas, we can make use of 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.
In this case, the heat added to the gas is given as 70.0 J. However, we need to convert the volume from cm3 to m3 as the SI unit for volume is m3.
Given:
Initial volume (V1) = 65.0 cm3 = 65.0 * 10^-6 m3
Final volume (V2) = 169.0 cm3 = 169.0 * 10^-6 m3
Pressure (P) = 1.7 atm
Heat added (Q) = 70.0 J
The work done by the gas can be calculated using the equation:
Work (W) = P * ∆V
Where ∆V is the change in volume, which can be calculated as:
∆V = V2 - V1
Let's calculate the work done by the gas:
∆V = (169.0 * 10^-6 m3) - (65.0 * 10^-6 m3) = 104.0 * 10^-6 m3
Work (W) = (1.7 atm) * (104.0 * 10^-6 m3) = 0.1768 J
Now, we can calculate the change in internal energy:
∆U = Q - W
∆U = 70.0 J - 0.1768 J = 69.8232 J
Therefore, the internal energy of the gas changed by approximately 69.8232 J.