One gram of water is placed in a cylinder and the pressure is maintained at 2.0 x 105 Pa. The temperature of the water is raised by 31oC. Determine the change in internal energy when, a) the water is in its liquid phase and causes an expansion of 1.0 x 10-8 m3. b) Water is in its gaseous phase and expands by the much greater amount of 7.1 x 10-5 m3. Assume that there is no phase change for both cases. (cliquidH20 =4186 J/(kgoC); cgasH20 = 2020 J/(kgoC))
To determine the change in internal energy, we need to use the equation:
ΔU = Q - W
where ΔU is the change in internal energy, Q is the heat added to or removed from the system, and W is the work done by or on the system.
a) For the water in its liquid phase, the equation becomes:
ΔU = Q - W
We can calculate the heat added using the formula:
Q = mcΔT
where m is the mass of the water, c is the specific heat capacity of water in the liquid phase, and ΔT is the change in temperature.
Given:
Mass of water (m) = 1 gram = 0.001 kg
Specific heat capacity of water in liquid phase (c liquid H2O) = 4186 J/(kg°C)
Change in temperature (ΔT) = 31°C
Substituting the values into the equation, we get:
Q = (0.001 kg) * (4186 J/(kg°C)) * (31°C)
= 129.506 J
Now, let's calculate the work done using the formula:
W = PΔV
where P is the pressure and ΔV is the change in volume.
Given:
Pressure (P) = 2.0 x 10^5 Pa
Change in volume (ΔV) = 1.0 x 10^(-8) m^3
Substituting the values into the equation, we get:
W = (2.0 x 10^5 Pa) * (1.0 x 10^(-8) m^3)
= 2.0 x 10^(-3) J
Now, substituting the values of Q and W into the equation for ΔU:
ΔU = 129.506 J - 2.0 x 10^(-3) J
= 129.504 J
Therefore, the change in internal energy when the water is in the liquid phase and causes an expansion of 1.0 x 10^(-8) m^3 is 129.504 J.
b) Now let's calculate the change in internal energy when the water is in its gaseous phase and expands by 7.1 x 10^(-5) m^3.
Using the same equation:
ΔU = Q - W
To calculate Q, we will again use the formula:
Q = mcΔT
We already have the values for mass (m) and change in temperature (ΔT) from before. However, the specific heat capacity for water in the gaseous phase (c gas H2O) is different and given as 2020 J/(kg°C).
Q = (0.001 kg) * (2020 J/(kg°C)) * (31°C)
≈ 62.86 J
Next, we calculate the work done (W):
W = PΔV
Given:
Pressure (P) = 2.0 x 10^5 Pa
Change in volume (ΔV) = 7.1 x 10^(-5) m^3
W = (2.0 x 10^5 Pa) * (7.1 x 10^(-5) m^3)
= 14.2 J
Now, substituting the values of Q and W into the equation for ΔU:
ΔU = 62.86 J - 14.2 J
≈ 48.66 J
Therefore, the change in internal energy when the water is in the gaseous phase and expands by 7.1 x 10^(-5) m^3 is approximately 48.66 J.