A cylinder fitted with a moveable piston contains 2L of oxygen at 25°C and 3 bar. The piston is pushed to compress the gas to a pressure of 6 bar. a. Write and simplify the energy balance for this system. (Do not assume the process is isothermal.) b. Now assume the process is isothermal. The compression work done on the gas is 4.6L-bar. Assume the gas is ideal (Û is a function only of T). How much heat is transferred to or from the system (be careful of the sign of Q)? Express Q in common heat units. c. Now assume the process is adiabatic. Is the final system temperature greater or less than 25°C. (Û increases as T increases.)

Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
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### Problem Statement:

A cylinder fitted with a moveable piston contains 2L of oxygen at 25°C and 3 bar. The piston is pushed to compress the gas to a pressure of 6 bar.

---

#### a. Write and simplify the energy balance for this system. (Do not assume the process is isothermal.)

#### b. Now assume the process is isothermal. The compression work done on the gas is 4.6L·bar. Assume the gas is ideal (Û is a function only of T). How much heat is transferred to or from the system (be careful of the sign of Q)? Express Q in common heat units.

#### c. Now assume the process is adiabatic. Is the final system temperature greater or less than 25°C. (Û increases as T increases.)

---

### Explanation:

#### a. Energy Balance (Non-Isothermal Process):
To write and simplify the energy balance for the system, you need to consider the first law of thermodynamics. The general energy balance equation for a control mass undergoing any process is:
\[ \Delta U = Q - W \]
where:
- \( \Delta U \) is the change in internal energy,
- \( Q \) is the heat added to the system,
- \( W \) is the work done by the system.

Since it is mentioned not to assume the process is isothermal, we will consider changes in internal energy with respect to temperature.

#### b. Isothermal Process:
For an isothermal process (constant temperature), the internal energy change \( \Delta U \) is zero for an ideal gas.

The work done on the gas is given as 4.6 L·bar. 

Using the first law of thermodynamics:
\[ \Delta U = Q - W \]
Since \( \Delta U = 0 \),
\[ 0 = Q - W \implies Q = W \]

Therefore, the heat transferred \( Q \) is equal to the work done \( W \):
\[ Q = 4.6 \, \text{L·bar} \]

To express \( Q \) in common heat units, recall that 1 L·bar = 100 J (joules).
\[ Q = 4.6 \times 100 = 460 \, \text{J} \]

#### c. Adiabatic Process:
For an adiabatic process, there is no heat exchange with the surroundings (\
Transcribed Image Text:### Problem Statement: A cylinder fitted with a moveable piston contains 2L of oxygen at 25°C and 3 bar. The piston is pushed to compress the gas to a pressure of 6 bar. --- #### a. Write and simplify the energy balance for this system. (Do not assume the process is isothermal.) #### b. Now assume the process is isothermal. The compression work done on the gas is 4.6L·bar. Assume the gas is ideal (Û is a function only of T). How much heat is transferred to or from the system (be careful of the sign of Q)? Express Q in common heat units. #### c. Now assume the process is adiabatic. Is the final system temperature greater or less than 25°C. (Û increases as T increases.) --- ### Explanation: #### a. Energy Balance (Non-Isothermal Process): To write and simplify the energy balance for the system, you need to consider the first law of thermodynamics. The general energy balance equation for a control mass undergoing any process is: \[ \Delta U = Q - W \] where: - \( \Delta U \) is the change in internal energy, - \( Q \) is the heat added to the system, - \( W \) is the work done by the system. Since it is mentioned not to assume the process is isothermal, we will consider changes in internal energy with respect to temperature. #### b. Isothermal Process: For an isothermal process (constant temperature), the internal energy change \( \Delta U \) is zero for an ideal gas. The work done on the gas is given as 4.6 L·bar. Using the first law of thermodynamics: \[ \Delta U = Q - W \] Since \( \Delta U = 0 \), \[ 0 = Q - W \implies Q = W \] Therefore, the heat transferred \( Q \) is equal to the work done \( W \): \[ Q = 4.6 \, \text{L·bar} \] To express \( Q \) in common heat units, recall that 1 L·bar = 100 J (joules). \[ Q = 4.6 \times 100 = 460 \, \text{J} \] #### c. Adiabatic Process: For an adiabatic process, there is no heat exchange with the surroundings (\
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