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 calculate the work done on the system and the heat transferred to the system.
a) When water is in its liquid phase and causes an expansion of 1.0 x 10^(-8) m^3:
The work done on the system can be calculated using the equation: W = P * ΔV, 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
Using the given values, we can calculate the work done on the system:
W = (2.0 x 10^5 Pa) * (1.0 x 10^(-8) m^3) = 2.0 J (approximately)
The heat transferred to the system can be calculated using the equation: Q = m * c * ΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Given:
Mass (m) = 1 gram = 0.001 kg
Specific heat capacity (c) = 4186 J/(kg·°C)
Change in temperature (ΔT) = 31°C
Using the given values, we can calculate the heat transferred to the system:
Q = (0.001 kg) * (4186 J/(kg·°C)) * (31°C) = 129.446 J (approximately)
The change in internal energy (ΔU) can be calculated using the first law of thermodynamics:
ΔU = Q - W
ΔU = 129.446 J - 2.0 J = 127.446 J (approximately)
Therefore, the change in internal energy when water is in its liquid phase and causes an expansion of 1.0 x 10^(-8) m^3 is approximately 127.446 J.
b) When water is in its gaseous phase and expands by the amount of 7.1 x 10^(-5) m^3:
The calculations for this scenario are similar to the previous case, but with a different change in volume.
Given:
Pressure (P) = 2.0 x 10^5 Pa
Change in volume (ΔV) = 7.1 x 10^(-5) m^3
Using the given values, we can calculate the work done on the system:
W = (2.0 x 10^5 Pa) * (7.1 x 10^(-5) m^3) = 14 J (approximately)
The heat transferred to the system can be calculated using the same equation:
Q = (0.001 kg) * (2020 J/(kg·°C)) * (31°C) = 62.62 J (approximately)
The change in internal energy can be calculated using the first law of thermodynamics:
ΔU = Q - W
ΔU = 62.62 J - 14 J = 48.62 J (approximately)
Therefore, the change in internal energy when water is in its gaseous phase and expands by the amount of 7.1 x 10^(-5) m^3 is approximately 48.62 J.