Derive the expression for heat exchanged in case of an isobaric process.
Answers
Answer:
Explanation:
The heat transferred to the system does work but also changes the internal energy of the system.
In an isobaric process for a monatomic gas , heat and temperature change satisfy the following equation : Q=5 2 N K ( TRIANGLE) T .
For a monatomic ideal gas, specific heat at constant pressure is 52R 5 2 R
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★Required Answer:★
➤ The initial volume and temperature of the gas is Vi, Ti and Final volume and temperature of gas is Vf, Tf and ∆p= 0 (pressure constant).
➙ The work done in the expansion is =W
➙ p dV = p (Vf - Vi) _________⑴
➤ From ideal gas equation, pV= nRT
➙ W = nR (Tf - Ti). __________⑵
The Change in internal energy is given by ∆U
∆U = nCv ∆T
➙ ∆U = nCv (Tf - Ti) _'_______⑶
➤ Cv is specific heat at constant Volume and ∆T is Change in temperature at constant pressure.
Applying First law of Thermodynamics to the isobaric process.
Q = ∆U + W
From equation ⑵ & ⑶
we get,
Q = nCv (Tf -Ti) + nR (Tf - Ti)
➙ Q = n Cv + nR (Tf - Ti) ____⑷
Q = n (Cv + R) (Tf - Ti)
∴ Q = n Cp (Tf - Ti) __________⑸
➙ ∴ (Cp - Cv = R)
★ By the above equation it can be concluded that since there is a temperature change there is a change in internal energy also.
★ The heat is used to increase the temperature as well as do work.
More to know:
★ Isobaric process:- ★
DEFINATION: A Thermodynamic process taking place at ᴄᴏɴsᴛᴀɴᴛ ᴘʀᴇssᴜʀᴇ is called an isobaric process.
✍︎ Important features of Isobaric process:-
- This is a constant pressure process.
- For such a system ∆p= 0 and ∆T ≠ 0
- In the process , a gas either expands or contracts to maintain constant pressure and hence the net amount of work is done by the system or on the system.
- For this process, the three quantities ∆U, Q and W are not equal to zero.
EXAMPLES :-
- An example of the isobaric process includes the boiling of water to steam or the freezing of water to ice.