An inflated rubber balloon contains one mole of an ideal gas has a pressure p, volume V and temperature T. if the temperature rises to 1.1 T, and the volume is increased to 1.05 V, the final pressure will be

Options

A. 1.1 p
B. p
C. less than p
D. between p and 1.1

Explanation

To solve the problem, we will use the Ideal Gas Law, which states that:

\[ PV = nRT \]

Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( n \) = Number of moles
- \( R \) = Ideal gas constant
- \( T \) = Temperature

Given:
- Initial conditions: \( P_1 = P \), \( V_1 = V \), \( T_1 = T \)
- Final conditions: \( T_2 = 1.1T \), \( V_2 = 1.05V \)

Since the number of moles \( n \) remains constant (1 mole in this case), we can set up the relationship using the Ideal Gas Law for the initial and final states:

1. **Set up the equation for initial state:**
\[ P_1 V_1 = n R T_1 \]
This can be written as:
\[ P V = 1 \cdot R T \]

2. **Set up the equation for final state:**
\[ P_2 V_2 = n R T_2 \]
This can be written as:
\[ P_2 (1.05V) = 1 \cdot R (1.1T) \]

3. **Equate the two states:**
From the initial state:
\[ P = \frac{R T}{V} \]
From the final state:
\[ P_2 = \frac{R (1.1T)}{1.05V} \]

4. **Substituting the expression for \( P \) into the equation for \( P_2 \):**
\[ P_2 = \frac{1.1T}{1.05V} \cdot \frac{R}{V} = P \cdot \frac{1.1}{1.05} \]

5. **Calculate \( P_2 \):**
\[ P_2 = P \cdot \frac{1.1}{1.05} \]
Simplifying this gives:
\[ P_2 = P \cdot 1.047619 \]
Approximately:
\[ P_2 \approx 1.05P \]

Thus, the final pressure \( P_2 \) is approximately \( 1.05P \).