Question

The specific heat of hydrogen gas at constant pressure ${C_p} = 3.4 \times {10^3}cal/kg{}^0C$ is and at constant volume is ${C_v} = 2.4 \times {10^3}cal/kg{}^0C$ If one kilogram hydrogen gas is heated from ${10^0}C$ to ${20^0}C$ at constant pressure, the external work done on the gas to maintain is at constant pressure is:
\[
  A.{\text{ }}{10^5}cal \\
  B.{\text{ }}{10^4}cal \\
  C.{\text{ }}{10^3}cal \\
  D.{\text{ }}5 \times {10^3}cal \\
 \]

Answer 1

Hint: In order to solve the given problem first we will see two different formulas, the first one relating the work done on the gas and the change in the energy of the gas. Further we will see the relation between the specific heat constant and the gas constant. We will further use the two formulas in combination one by one to find out some middle terms and finally the external work done on the gas in the given case.

Complete step by step answer:
From the first law of thermodynamics we know the relation that:
$\Delta Q = \Delta U + \Delta W$
Here the term $\Delta U$ represents change in the internal energy, the term $\Delta Q$ represents heat added or the energy added and the term $\Delta W$ represents work done.

From the above formula work done at constant pressure is:
${\left( {\Delta W} \right)_P} = {\left( {\Delta Q} \right)_P} - \Delta U$
For the given case of constant pressure we know that the change in the internal energy is the same as the heat added at the constant volume. So the formula becomes.
${\left( {\Delta W} \right)_P} = {\left( {\Delta Q} \right)_P} - {\left( {\Delta Q} \right)_V}$ ----- (1)

For the given problem we have:
Specific heat at constant pressure ${C_p} = 3.4 \times {10^3}cal/kg{}^0C$
Specific heat at constant pressure ${C_v} = 2.4 \times {10^3}cal/kg{}^0C$
Also we know that the mass of hydrogen gas is 1 kg.
We know that the initial temperature is ${10^0}C = \left( {10 + 273} \right)K = 283K$ .
And the final temperature is ${20^0}C = \left( {0 + 273} \right)K = 293K$ .
And we need to find out the work done.

As we know that the relation between the energy at constant pressure and volume with specific heat constant is:
$
  {\left( {\Delta Q} \right)_P} = m{C_P}\Delta T - - - - (2) \\
  {\left( {\Delta Q} \right)_V} = m{C_V}\Delta T - - - - (3) \\
 $
Let us substitute equation (2) and equation (3) in equation (1).
So we get the new formula for work at constant pressure is:
$
  \because {\left( {\Delta W} \right)_P} = {\left( {\Delta Q} \right)_P} - {\left( {\Delta Q} \right)_V} \\
   \Rightarrow {\left( {\Delta W} \right)_P} = m{C_P}\Delta T - m{C_V}\Delta T \\
   \Rightarrow {\left( {\Delta W} \right)_P} = m\left( {{C_P} - {C_V}} \right)\Delta T \\
 $
Now let us substitute the values in the above equation to find the value of work done.
$
  \because {\left( {\Delta W} \right)_P} = m\left( {{C_P} - {C_V}} \right)\Delta T \\
   \Rightarrow {\left( {\Delta W} \right)_P} = 1 \times \left( {3.4 \times {{10}^3} - 2.4 \times {{10}^3}} \right)\left( {293 - 283} \right) \\
   \Rightarrow {\left( {\Delta W} \right)_P} = 1 \times \left( {1 \times {{10}^3}} \right)\left( {10} \right) \\
   \Rightarrow {\left( {\Delta W} \right)_P} = {10^4}cal \\
 $
Hence, the external work done on the gas to maintain it at constant pressure is ${10^4}cal$ .
So, the correct answer is “Option B”.

Note: In order to solve such types of problems students must remember different formulas related to thermodynamics and must know the methods to relate them according to the requirement. Students must take special care of the units used. In such types of calculations we always take the temperature in Kelvin scale and not in Celsius scale. So students must convert the unit of temperature if it is different before solving the problem.