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Consider a 3-mole sample of an ideal monatomic gas initially at a temperature of 200 K and a volume of 10 liters. The gas undergoes an isochoric process, during which it absorbs 1500 J of heat. If the final temperature of the gas is 400 K, calculate the heat capacity at constant volume \mathrm{(C_v)} for this gas.

Given: Initial temperature \mathrm{(T_1)} = 200 K
Initial volume \mathrm{(V_1)} = 10 liters
Heat absorbed (Q) = 1500 J
Final temperature \mathrm{(V_2)} = 400 K
The gas is ideal and monatomic.

Option: 1

\mathrm{C_v}=2.50 \mathrm{~J} / \mathrm{K} \\


Option: 2

\mathrm{C_v}=3.00 \mathrm{~J} / \mathrm{K} \\


Option: 3

\mathrm{C_v}=7.50 \mathrm{~J} / \mathrm{K} \\


Option: 4

\mathrm{C_v}=10.00 \mathrm{~J} / \mathrm{K}


Answers (1)

best_answer

The heat capacity at constant volume \left ( \mathrm{C_{V}} \right ) is defined as the amount of heat required to raise the temperature of a substance by 1 degree Celsius (or 1 Kelvin) while keeping the volume constant. It is given by the equation:

                                         \mathrm{C_v=\frac{Q}{\Delta T}}

where: \mathrm{C_v} is the heat capacity at constant volume, Q is the heat absorbed by the gas during the process, and \mathrm{\Delta T} is the change in temperature (final temperature minus initial temperature).

The change in temperature \left ( \mathrm{\Delta T} \right ) is calculated as:

                               \mathrm{\Delta T=T_2-T_1=400 \mathrm{~K}-200 \mathrm{~K}=200 \mathrm{~K}}

Now, let’s calculate the heat capacity at constant volume \left ( \mathrm{C_{V}} \right ) using the formula:

                                \mathrm{C_v=\frac{Q}{\Delta T}}

Substitute the given values: Q = 1500 J
\mathrm{\Delta T = 200 K}
Calculating the value of \mathrm{C_v} :

                                    \mathrm{C_v=\frac{1500 \mathrm{~J}}{200 \mathrm{~K}} \approx 7.50 \mathrm{~J} / \mathrm{K}}

Answer:The heat capacity at constant volume \left ( \mathrm{C_{V}} \right ) for the given monatomic ideal gas is approximately 7.50 J/K.
So, the correct option is (3)

Posted by

vishal kumar

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