Question

10 mole of oxygen is heated at constant volume from 30 \(^{\circ}\)C to 40 \(^{\circ}\)C. The change in the internal energy of the gas is_____ cal.
(The molecular specific heat of oxygen at constant pressure, \(C_p=7\) cal/mol \(^{\circ}\)C and \(R=2\) cal/mol \(^{\circ}\)C.)

Hint:

\(\Delta U = nC_v \Delta T\) is valid for ideal gases in ANY process, not just isochoric. For isochoric process, \(Q = \Delta U\).

Answer 1

Correct Answer: 500

To solve the problem, we need to calculate the change in internal energy (\( \Delta U \)) of oxygen when heated at constant volume. For a given gas undergoing a temperature change at constant volume, the change in internal energy can be calculated using the formula:
\[ \Delta U = nC_v\Delta T \]
where \( n \) is the number of moles, \( C_v \) is the specific heat at constant volume, and \( \Delta T \) is the change in temperature.

First, we need to determine \( C_v \). Using the relation between specific heats, \( C_p - C_v = R \), we find:
\[ C_v = C_p - R \]
Given \( C_p = 7 \) cal/mol \(^{\circ}\)C and \( R = 2 \) cal/mol \(^{\circ}\)C, substituting these values gives:
\[ C_v = 7 - 2 = 5 \text{ cal/mol }^{\circ}\text{C} \]

Next, calculate \( \Delta T \):
\[ \Delta T = 40^{\circ}\text{C} - 30^{\circ}\text{C} = 10^{\circ}\text{C} \]

Now, substitute the values into the formula for \( \Delta U \):
\[ \Delta U = (10 \text{ moles})(5 \text{ cal/mol }^{\circ}\text{C})(10^{\circ}\text{C}) \]
\[ \Delta U = 500 \text{ cal} \]

The change in internal energy of the gas is \( 500 \) cal. This result is within the expected range of [500,500] cal, confirming its correctness.