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 with \( \Delta T = 50^\circ \text{C} \).

- Heat added is \( Q = n C_p \Delta T = E_1 \).

- Change in internal energy is \( \Delta U = n C_v \Delta T = E_2 \).

Step 1: Relate energy ratio to heat capacity ratio

The ratio \( \frac{E_1}{E_2} \) equals the ratio \( \frac{C_p}{C_v} \), which is \( \gamma \).

Step 2: Determine \( \gamma \) for a monoatomic gas

For a monoatomic gas with \( f=3 \) degrees of freedom, \( \gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3} \).

Step 3: Solve for \( x \)

Given the equation \( \frac{5}{3} = \frac{x}{9} \), solving for \( x \) yields \( x = 15 \).

Final Answer:

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