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 first analyze the two processes: the first one where the piston is fixed (constant volume) and the second one which is an isobaric process (constant pressure).

### Step 1: Understand the first process (constant volume)
In the first process, the piston is fixed, meaning the volume of the gas does not change. When heat \( \Delta Q \) is added to the gas, the change in internal energy \( \Delta U \) is equal to the heat added, since no work is done (work done \( W = 0 \)):
\[ \Delta Q = \Delta U \]

For an ideal monoatomic gas, the change in internal energy can be expressed as:
\[ \Delta U = N C_V \Delta T \]
where \( C_V \) is the molar heat capacity at constant volume and \( \Delta T \) is the change in temperature.

Given that the initial temperature is \( 400 \, K \) and the temperature increases by \( 20 \, K \):
\[ \Delta T = 20 \, K \]
Thus, we can write:
\[ \Delta Q = N C_V (20) \]

### Step 2: Identify the heat capacity for monoatomic gas
For a monoatomic ideal gas, the molar heat capacities are:
\[ C_V = \frac{3}{2} R \quad \text{and} \quad C_P = \frac{5}{2} R \]
where \( R \) is the universal gas constant.

### Step 3: Understand the second process (isobaric)
In the second process, heat \( \Delta Q \) is added at constant pressure. The relationship for heat added at constant pressure is:
\[ \Delta Q = N C_P \Delta T' \]
where \( \Delta T' \) is the change in temperature in the isobaric process.

### Step 4: Equate the heat added in both processes
Since the same amount of heat \( \Delta Q \) is added in both processes, we can set the two equations equal:
\[ N C_V (20) = N C_P \Delta T' \]
The \( N \) cancels out:
\[ C_V (20) = C_P \Delta T' \]

### Step 5: Substitute the values of \( C_V \) and \( C_P \)
Now substituting \( C_V \) and \( C_P \):
\[ \frac{3}{2} R (20) = \frac{5}{2} R \Delta T' \]
The \( R \) cancels out:
\[ \frac{3}{2} (20) = \frac{5}{2} \Delta T' \]

### Step 6: Solve for \( \Delta T' \)
Now we can solve for \( \Delta T' \):
\[ 30 = 5 \Delta T' \]
\[ \Delta T' = \frac{30}{5} = 6 \, K \]

### Final Answer
The change in temperature in the second process (isobaric) is \( \Delta T' = 6 \, K \).

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