Question

The temperature of 1 mole of an ideal monoatomic gas is increased by \( 50^\circ {C} \) at constant pressure. The total heat added and change in internal energy are \( E_1 \) and \( E_2 \), respectively. If \( \frac{E_1}{E_2} = \frac{x}{9} \), then the value of \( x \) is:

Hint:

The heat added at constant pressure and the change in internal energy are related to the specific heat capacities \( C_p \) and \( C_v \), respectively. For a monoatomic ideal gas, \( C_p = \frac{5}{2} R \) and \( C_v = \frac{3}{2} R \).

Answer 1

Given:

- The process is isobaric, and \( \Delta T = 50^\circ \text{C} \). - The heat added in an isobaric process is \( Q = n C_p \Delta T = E_1 \). - The change in internal energy in an isobaric process is \( \Delta U = n C_v \Delta T = E_2 \).

Step 1: Relating the energies and the heat capacities

Since \( \frac{E_1}{E_2} = \frac{C_p}{C_v} = \gamma \), we can relate the ratio of the heat capacities to the ratio of the energies. \[ \frac{E_1}{E_2} = \frac{C_p}{C_v} = \gamma. \]

Step 2: Use the specific heat capacities for a monoatomic gas

For a monoatomic gas, the value of \( \gamma \) is given by: \[ \gamma = 1 + \frac{2}{f}, \] where \( f \) is the number of degrees of freedom of the gas. For a monoatomic gas, \( f = 3 \). Substituting this value: \[ \gamma = 1 + \frac{2}{3} = \frac{5}{3}. \]

Step 3: Solving for \( x \)

The equation given in the problem is: \[ \frac{5}{3} = \frac{x}{9}. \] Solving for \( x \): \[ x = 15. \]

Final Answer:

The value of \( x \) is \( \boxed{15} \).