One mole of a perfect gas expands adiabatically .As a result of this,its pressure ,temperature and volume changes from P1,T1,V1, to P2,T2,V2 respectively .If molar specific heat at volume is Cv then work done by the gas wil be ?

Answer

To solve the problem of finding the work done by one mole of a perfect gas during an adiabatic expansion, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Adiabatic Process**:
In an adiabatic process, there is no heat exchange with the surroundings. Therefore, the heat transfer \( dq \) is equal to zero.

2. **Apply the First Law of Thermodynamics**:
The first law of thermodynamics states:
\[ dq = du + dw \]
Since \( dq = 0 \) for an adiabatic process, we can rewrite this as:
\[ 0 = du + dw \]
This implies:
\[ du = -dw \]

3. **Relate Change in Internal Energy to Temperature**:
The change in internal energy \( du \) for one mole of gas can be expressed using the specific heat at constant volume \( C_v \):
\[ du = C_v dT \]
where \( dT \) is the change in temperature.

4. **Substitute for \( du \) in the First Law Equation**:
From the previous steps, we can substitute \( du \) in the first law equation:
\[ C_v dT = -dw \]
Rearranging gives us:
\[ dw = -C_v dT \]

5. **Integrate to Find Work Done**:
To find the total work done \( W \) during the expansion from an initial temperature \( T_1 \) to a final temperature \( T_2 \), we integrate:
\[ W = \int_{T_1}^{T_2} -C_v dT = -C_v \int_{T_1}^{T_2} dT \]
This simplifies to:
\[ W = -C_v (T_2 - T_1) \]
Therefore, the work done by the gas is:
\[ W = C_v (T_1 - T_2) \]

### Final Result:
The work done by the gas during the adiabatic expansion is:
\[ W = C_v (T_1 - T_2) \]