Question

The 40 g of Argon is heated from \[{\rm{4}}{{\rm{0}}^{\rm{0}}}{\rm{C}}\]to \[{\rm{10}}{{\rm{0}}^{\rm{0}}}{\rm{C}}\] \[\left( {R = 2cal/mole} \right)\]. Then how much heat is absorbed at constant volume?
A. 100 cal
B. 80 cal
C. 180 cal
D. 120 cal

Answer 1

Hint:Before we proceed with the problem let’s see what is given. The number of moles of argon and change in temperature are given. We need to find the heat absorbed at a constant volume. For that first, we need to find the value of the amount of moles in argon thereby finding the heat absorbed at constant volume by the below formula. The molar mass is defined as the sum of the total mass in grams of the atoms present to make up a molecule per mole and its unit of molar is grams/mole.

Formula Used:
The formula to find the heat absorbed at the constant volume is given by,
\[Q = n{C_V}dT\]………… (1)
Where,
\[n\] is the amount of moles of substance.
\[{C_V}\] is specific heat at constant volume.
[\ dT\] is a change in temperature.

Complete step by step solution:
The heat absorbed at the constant volume is given by,
\[Q = n{C_V}dT\]
Given that argon consists of 40g that is \[m = 40g\] and we need to find the value of n. To find n we have a formula from the molar mass that is,
\[M = \dfrac{m}{n}\]
\[\Rightarrow n = \dfrac{m}{M}\]
Here, m is the mass of argon in grams and M is the molar mass of argon and its standard value is 40u.
\[ n = \dfrac{{40}}{{40}} = 1\]

Since it is a monatomic gas, \[{C_V} = \dfrac{3}{2}R\]
where the value of \[\left( {R = 2cal/mole} \right)\] and \[dT = {(100 - 40)^0}C\]
Then, equation (1) becomes
\[Q = 1 \times \dfrac{3}{2} \times 2 \times \left( {100 - 40} \right)\]
\[\Rightarrow Q = 3 \times \left( {60} \right)\]
\[ \Rightarrow Q = 180\,cal\]
Therefore, the heat is absorbed at a constant volume of 180 cal.

Hence, option C is the correct answer.

Note:Specific heat at constant volume is defined as the quantity of heat which is required to raise the temperature of unit mass of gas by 1 degree at constant volume and is denoted by \[{C_V}\].