Two moles of an ideal gas at 1 atm are compressed to 2 atm at 273 K. The enthalpy change for the process is

Options

A. 2 litre atm
B. 1 litre atm
C. zero
D. 1 litre atm

Explanation

To solve the problem of finding the enthalpy change for the compression of an ideal gas, we can follow these steps:

### Step 1: Understand the properties of an ideal gas
An ideal gas follows the ideal gas law, and its behavior can be described by the equation:
\[ PV = nRT \]
where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature.

### Step 2: Identify the conditions of the gas
We are given:
- Initial pressure \( P_1 = 1 \, \text{atm} \)
- Final pressure \( P_2 = 2 \, \text{atm} \)
- Number of moles \( n = 2 \)
- Temperature \( T = 273 \, \text{K} \)

### Step 3: Recall the enthalpy change for an ideal gas
For an ideal gas, the enthalpy \( H \) is a function of temperature only. Therefore, if the temperature remains constant during the process (isothermal process), the enthalpy change \( \Delta H \) can be expressed as:
\[ \Delta H = nC_p\Delta T \]
where \( C_p \) is the molar heat capacity at constant pressure and \( \Delta T \) is the change in temperature.

### Step 4: Determine if there is a change in temperature
Since the problem states that the gas is compressed at a constant temperature of 273 K, we have:
\[ \Delta T = 0 \]

### Step 5: Calculate the enthalpy change
Substituting \( \Delta T = 0 \) into the enthalpy change equation:
\[ \Delta H = nC_p \cdot 0 = 0 \]

Thus, the enthalpy change for the process is:
\[ \Delta H = 0 \]

### Conclusion
The enthalpy change for the compression of the ideal gas is \( 0 \, \text{J} \).