The pressure exerted by `12 g` of an ideal gas at temperature `t^(@)C` in a vessel of volume `V litre` is `1 atm`. When the temperature is increased by `10^(@)C` at the same volume, the pressure increases by `10%`. Calculate the temperature `t` and volume `V`. (Molecular weight of the gas is `120`).

Explanation

To solve the problem step-by-step, we will use the ideal gas law and the relationship between pressure and temperature.

### Step 1: Understand the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 2: Convert Temperature to Kelvin
The temperature in Celsius can be converted to Kelvin using the formula:
\[ T(K) = t(°C) + 273.15 \]

### Step 3: Determine Initial Conditions
From the problem:
- Initial pressure \( P_1 = 1 \, \text{atm} \)
- Initial temperature \( T_1 = t + 273.15 \, \text{K} \)
- Mass of the gas \( m = 12 \, \text{g} \)
- Molar mass \( M = 120 \, \text{g/mol} \)

### Step 4: Calculate Moles of Gas
The number of moles \( n \) can be calculated using:
\[ n = \frac{m}{M} = \frac{12 \, \text{g}}{120 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 5: Set Up the Equation for Pressure Change
When the temperature is increased by \( 10°C \), the new temperature \( T_2 \) becomes:
\[ T_2 = (t + 10) + 273.15 = t + 283.15 \, \text{K} \]

The new pressure \( P_2 \) after the temperature increase is:
\[ P_2 = P_1 + 0.1 P_1 = 1.1 \, \text{atm} \]

### Step 6: Use the Ideal Gas Law for Both Conditions
Using the ideal gas law for both states:
1. For initial conditions:
\[ P_1 V = n R T_1 \]
2. For final conditions:
\[ P_2 V = n R T_2 \]

### Step 7: Set Up the Ratio of Pressures and Temperatures
From the ideal gas law, we can set up the ratio:
\[ \frac{P_1}{P_2} = \frac{T_1}{T_2} \]
Substituting the known values:
\[ \frac{1}{1.1} = \frac{t + 273.15}{t + 283.15} \]

### Step 8: Cross Multiply and Solve for \( t \)
Cross multiplying gives:
\[ 1.1(t + 273.15) = (t + 283.15) \]
Expanding and rearranging:
\[ 1.1t + 300.465 = t + 283.15 \]
\[ 0.1t = 283.15 - 300.465 \]
\[ 0.1t = -17.315 \]
\[ t = -173.15 \, °C \]

### Step 9: Calculate Volume \( V \)
Using the ideal gas law to find volume:
\[ V = \frac{nRT}{P} \]
Substituting the values:
- \( n = 0.1 \, \text{mol} \)
- \( R = 0.0821 \, \text{L atm/(mol K)} \)
- \( T = 100 \, \text{K} \) (since \( t + 273.15 = 100 \))

Calculating volume:
\[ V = \frac{0.1 \times 0.0821 \times 100}{1} = 0.821 \, \text{L} \]

### Final Answers
- Temperature \( t = -173.15 \, °C \)
- Volume \( V = 0.82 \, \text{L} \)