A cylinder fitted with a piston contains an ideal monoatomic gas at a temperature of `400K`. The piston is held fixed while heat `DeltaQ` is given to the gas, it is found the temperature of the gas has increased by `20K`. In an isobaric process the same `DeltaQ` heat is supplied slowely to it. Find the change in temperature in the second process?

Answer

To solve the problem step by step, we will analyze the two processes: the isochoric process and the isobaric process.

### Step 1: Understand the Isochoric Process
In the first process, the gas is heated at constant volume (isochoric process). The heat added to the system is given by the equation:
\[ \Delta Q = \Delta U \]
where \(\Delta U\) is the change in internal energy. For an ideal monoatomic gas, the change in internal energy can be expressed as:
\[ \Delta U = n C_v \Delta T \]
Here, \(C_v\) is the specific heat at constant volume, and \(\Delta T\) is the change in temperature. From the problem, we know that \(\Delta T = 20 \, K\).

### Step 2: Relate Heat to Change in Temperature
Substituting the known values into the equation for the isochoric process:
\[ \Delta Q = n C_v (20) \]

### Step 3: Understand the Isobaric Process
In the second process, the gas is heated at constant pressure (isobaric process). The heat added to the system in this case is given by:
\[ \Delta Q = n C_p \Delta T' \]
where \(\Delta T'\) is the change in temperature during the isobaric process and \(C_p\) is the specific heat at constant pressure.

### Step 4: Equate the Heat for Both Processes
Since the same amount of heat \(\Delta Q\) is supplied in both processes, we can equate the two expressions for heat:
\[ n C_v (20) = n C_p \Delta T' \]
We can cancel \(n\) from both sides (assuming \(n \neq 0\)):
\[ C_v (20) = C_p \Delta T' \]

### Step 5: Use the Relation Between \(C_p\) and \(C_v\)
For a monoatomic ideal gas, the relation between \(C_p\) and \(C_v\) is given by:
\[ \frac{C_p}{C_v} = \frac{5}{3} \]
This implies:
\[ C_p = \frac{5}{3} C_v \]

### Step 6: Substitute \(C_p\) in the Heat Equation
Substituting \(C_p\) into the equation:
\[ C_v (20) = \left(\frac{5}{3} C_v\right) \Delta T' \]
We can cancel \(C_v\) from both sides (assuming \(C_v \neq 0\)):
\[ 20 = \frac{5}{3} \Delta T' \]

### Step 7: Solve for \(\Delta T'\)
To find \(\Delta T'\), we rearrange the equation:
\[ \Delta T' = 20 \cdot \frac{3}{5} = 12 \, K \]

### Final Answer
The change in temperature in the second process (isobaric) is:
\[ \Delta T' = 12 \, K \]