Two moles of a gas at temperature T and volume V are heated to twice its volume at constant pressure. If `(C _(p))/(C _(v)) = gamma` then increase in internal energy of the gas is-

Options

A. `(RT)/(gamma - 1)`
B. `(2 RT)/( gamma - 1)`
C. `(2 RT)/( 3( gamma - 1))`
D. `(2T)/( gamma - 1)`

Answer

To find the increase in internal energy of the gas when it is heated at constant pressure, we can follow these steps:

### Step 1: Understand the relationship between internal energy, temperature, and heat capacities.
The change in internal energy (\( \Delta U \)) for an ideal gas can be expressed as:
\[ \Delta U = n C_v \Delta T \]
where \( n \) is the number of moles, \( C_v \) is the molar heat capacity at constant volume, and \( \Delta T \) is the change in temperature.

### Step 2: Determine the change in temperature (\( \Delta T \)).
Since the gas is heated at constant pressure and its volume is doubled, we can use the ideal gas law:
\[ PV = nRT \]
Given that the pressure is constant, if the volume doubles, the temperature must also double. Therefore, if the initial temperature is \( T \), the final temperature \( T_f \) will be:
\[ T_f = 2T \]
Thus, the change in temperature is:
\[ \Delta T = T_f - T = 2T - T = T \]

### Step 3: Relate \( C_v \) to \( C_p \) using the given ratio \( \frac{C_p}{C_v} = \gamma \).
From the relationship between heat capacities:
\[ C_p = \gamma C_v \]
We can also use the relation:
\[ C_p - C_v = R \]
Substituting \( C_p \):
\[ \gamma C_v - C_v = R \]
This simplifies to:
\[ C_v(\gamma - 1) = R \]
Thus, we can express \( C_v \):
\[ C_v = \frac{R}{\gamma - 1} \]

### Step 4: Substitute \( C_v \) and \( \Delta T \) into the equation for \( \Delta U \).
Now substituting \( C_v \) and \( \Delta T \) into the equation for internal energy:
\[ \Delta U = n C_v \Delta T = 2 \left(\frac{R}{\gamma - 1}\right) T \]
This gives us:
\[ \Delta U = \frac{2RT}{\gamma - 1} \]

### Final Answer:
The increase in internal energy of the gas is:
\[ \Delta U = \frac{2RT}{\gamma - 1} \]