Question

A certain amount of heat energy is supplied to a monoatomic ideal gas which expands at constant pressure. What fraction of the heat energy is converted into work ?

• A. 1
• B. $\frac {2}{3}$
• C. $\frac {2}{5}$
• D. $\frac {5}{7}$

Answer 1

The Correct Option is C

When a monoatomic ideal gas expands at constant pressure, part of the supplied heat energy is used to do work and the rest increases the internal energy of the gas. For an ideal gas expanding at constant pressure, the heat energy supplied is given by the formula: \[ Q = \Delta U + W \] Where:
\( Q \) is the total heat energy supplied, 
\( \Delta U \) is the change in internal energy,
\( W \) is the work done by the gas.
For a monoatomic ideal gas, the change in internal energy is given by:
\[ \Delta U = \frac{3}{2} n R \Delta T \] The work done by the gas during expansion at constant pressure is: \[ W = P \Delta V = n R \Delta T \] The fraction of heat energy converted into work is: \[ \frac{W}{Q} = \frac{n R \Delta T}{\frac{5}{2} n R \Delta T} = \frac{2}{5} \] Thus, the fraction of heat energy converted into work is \(\frac{2}{5}\).

Answer 2

The heat $Q$ is converted into the internal energy and work. According to the first law of thermodynamics,
$Q = U + W$
$ \Rightarrow nC _{ p } \Delta t = n C _{ v } \Delta t + W$
$\frac{W}{Q}=\frac{n C_{p} \Delta t-n C_{v} \Delta t}{n C_{p} \Delta t} $
$\frac{c_{p}-c_{v}}{c_{p}}=1-\frac{1}{\gamma}$
$=1-\frac{1}{\frac{5}{3}}=\frac{2}{5}$

Answer 3

For an ideal gas expanding at constant pressure, the first law of thermodynamics gives us: \[ \Delta Q = \Delta U + W \] where: - \( \Delta Q \) is the heat energy supplied, - \( \Delta U \) is the change in internal energy, - \( W \) is the work done by the gas. For a monoatomic ideal gas, the change in internal energy \( \Delta U \) is related to the change in temperature \( \Delta T \) as: \[ \Delta U = \frac{3}{2} n R \Delta T \] where: - \( n \) is the number of moles, - \( R \) is the gas constant. The work done by the gas in an isobaric (constant pressure) expansion is given by: \[ W = P \Delta V = n R \Delta T \] where \( P \Delta V \) is the work done. Now, the heat energy supplied \( \Delta Q \) is: \[ \Delta Q = \Delta U + W = \frac{3}{2} n R \Delta T + n R \Delta T = \frac{5}{2} n R \Delta T \] The fraction of the heat energy converted into work is: \[ \frac{W}{\Delta Q} = \frac{n R \Delta T}{\frac{5}{2} n R \Delta T} = \frac{2}{5} \]

Thus, the fraction of heat energy converted into work is \({\frac{2}{5}} \).