JEE Mains · Physics · STD 11 - Kinetic theory of gases
- JEE Mains 2026
- Medium
- Numerical
\(5\) moles of unknown gas is heated at constant volume from \(10 \, °C\) to \(20 \, °C\). The molar specific heat of this gas at constant pressure \(c_p = 8\) cal/mol.°C and \(R = 8.36\) J/mol.°C. The change in the internal energy of the gas is _______ calorie.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) 300
Step-by-step Solution
Detailed explanation
Given \(n = 5\) moles, \(\Delta T = 20^{\circ}\text{C} - 10^{\circ}\text{C} = 10^{\circ}\text{C}\).
The molar specific heat at constant pressure is \(c_p = 8 \text{ cal/mol}\cdot^{\circ}\text{C}\).
The universal gas constant is \(R = 8.36 \text{ J/mol}\cdot^{\circ}\text{C}\).
Converting \(R\) into calories, we use \(1 \text{ cal} = 4.18 \text{ J}\):
\(R = \dfrac{8.36}{4.18} \text{ cal/mol}\cdot^{\circ}\text{C} = 2 \text{ cal/mol}\cdot^{\circ}\text{C}\)
Using Mayer's relation, the molar specific heat at constant volume is:
\(c_v = c_p - R = 8 - 2 = 6 \text{ cal/mol}\cdot^{\circ}\text{C}\)
The change in internal energy of the gas is given by:
\(\Delta U = n c_v \Delta T\)
\(\Delta U = 5 \times 6 \times 10 = 300 \text{ cal}\)
Answer: \(300\)
The molar specific heat at constant pressure is \(c_p = 8 \text{ cal/mol}\cdot^{\circ}\text{C}\).
The universal gas constant is \(R = 8.36 \text{ J/mol}\cdot^{\circ}\text{C}\).
Converting \(R\) into calories, we use \(1 \text{ cal} = 4.18 \text{ J}\):
\(R = \dfrac{8.36}{4.18} \text{ cal/mol}\cdot^{\circ}\text{C} = 2 \text{ cal/mol}\cdot^{\circ}\text{C}\)
Using Mayer's relation, the molar specific heat at constant volume is:
\(c_v = c_p - R = 8 - 2 = 6 \text{ cal/mol}\cdot^{\circ}\text{C}\)
The change in internal energy of the gas is given by:
\(\Delta U = n c_v \Delta T\)
\(\Delta U = 5 \times 6 \times 10 = 300 \text{ cal}\)
Answer: \(300\)
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