A perfect gas contained in a cylinder is kept in vacuum. If the cylinder suddenly bursts, then the temperature of the gas

Options

A. (a)remains constant
B. (b)becomes zero
C. (c)increases
D. (d)decreases

Answer

To solve the problem of what happens to the temperature of a perfect gas when a cylinder containing it bursts in a vacuum, we can follow these steps:

### Step 1: Understand the Conditions
The gas is contained in a cylinder that is kept in a vacuum. When the cylinder bursts, the gas expands freely into the vacuum.

**Hint:** Remember that in a vacuum, there is no external pressure exerted on the gas.

### Step 2: Analyze Free Expansion
When the gas expands freely (i.e., without doing work against an external pressure), the work done (W) is zero. This is a key characteristic of free expansion.

**Hint:** Free expansion means the gas does not push against anything, so no work is done.

### Step 3: Consider Heat Exchange
Since the gas is in a vacuum, there is no heat exchange with the surroundings. This means that the heat transfer (Q) is also zero.

**Hint:** In a vacuum, there are no particles to transfer heat, so Q = 0.

### Step 4: Apply the First Law of Thermodynamics
According to the first law of thermodynamics, the change in internal energy (ΔU) is given by the equation:
\[ \Delta U = Q - W \]
Since both Q and W are zero in this case:
\[ \Delta U = 0 - 0 = 0 \]

**Hint:** The first law of thermodynamics relates internal energy, heat, and work.

### Step 5: Relate Internal Energy to Temperature
For an ideal gas, the internal energy is directly proportional to the temperature. Therefore, if there is no change in internal energy (ΔU = 0), it implies that there is no change in temperature (ΔT = 0).

**Hint:** Remember that for an ideal gas, internal energy is a function of temperature.

### Conclusion
Since there is no change in internal energy, the temperature of the gas remains constant after the cylinder bursts.

**Final Answer:** The temperature of the gas will remain constant.