The volume of one mole of ideal gas with adiabatic exponent is varied according to law `V = 1/T`. Find amount of heat obtained by gas in this process if gas temperature is increased by 100 K.

Options

A. `100 R`
B. `200 R`
C. `300 R`
D. `400 R`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between volume and temperature
The volume of the gas is given by the equation \( V = \frac{1}{T} \). This implies that \( V \cdot T = 1 \).

### Step 2: Use the ideal gas equation
The ideal gas equation is given by:
\[ PV = nRT \]
For one mole of gas (\( n = 1 \)), we can express temperature \( T \) in terms of pressure \( P \) and volume \( V \):
\[ T = \frac{PV}{R} \]

### Step 3: Substitute the expression for \( T \) into the volume equation
Substituting \( T \) from the ideal gas equation into the volume equation:
\[ V = \frac{1}{T} = \frac{R}{PV} \]
Multiplying both sides by \( PV \):
\[ PV^2 = R \]
This shows that \( PV^2 \) is a constant.

### Step 4: Identify the adiabatic exponent
In an adiabatic process, we have the relation:
\[ PV^x = \text{constant} \]
From our previous step, we see that \( x = 2 \).

### Step 5: Calculate the molar specific heat \( C \)
For an adiabatic process, the molar specific heat can be calculated using:
\[ C = C_v + \frac{R}{1 - x} \]
Where \( C_v \) is the molar specific heat at constant volume. For a monatomic ideal gas, \( C_v = \frac{3}{2}R \). Thus:
\[ C = \frac{3}{2}R + \frac{R}{1 - 2} \]
Calculating this gives:
\[ C = \frac{3}{2}R - R = \frac{1}{2}R \]

### Step 6: Calculate the heat obtained by the gas
The heat obtained by the gas can be calculated using the formula:
\[ q = nC\Delta T \]
Substituting the values:
- \( n = 1 \) (one mole of gas)
- \( C = \frac{1}{2}R \)
- \( \Delta T = 100 \, K \)

Thus:
\[ q = 1 \cdot \frac{1}{2}R \cdot 100 = 50R \]

### Final Answer
The amount of heat obtained by the gas in this process is \( 50R \).

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