Question 714: Thermodynamics & KTG

Question

A monoatomic ideal gas is heated at constant pressure. The percentage of total heat used in increasing the internal energy and that used for doing external work is $A$ and $B$ respectively. Then the ratio, $\mathrm{A}: \mathrm{B}$ is

Accepted Answer

Let us solve step by step:
Step 1: Total Heat Supplied at Constant Pressure
For $n$ moles of an ideal gas heated at constant pressure,
Heat supplied: $Q = nC_p \Delta T$
Where $C_p$ is molar specific heat at constant pressure.
For a
monoatomic ideal gas
:
$C_v = \frac{3}{2}R$
$C_p = C_v + R = \frac{5}{2}R$
So,
$$Q = nC_p \Delta T = n \left(\frac{5}{2}R\right) \Delta T$$
Step 2: Increase in Internal Energy $(\Delta U)$
For $n$ moles,
$$\Delta U = nC_v \Delta T = n \left( \frac{3}{2} R \right) \Delta T$$
Step 3: Work Done by the Gas $(W)$ at Constant Pressure
$$W = P \Delta V = nR \Delta T$$
Step 4: Find Percentage of Heat Used for $\Delta U$ and $W$
Fraction for $\Delta U$:
$$\frac{\Delta U}{Q} = \frac{n \cdot \frac{3}{2}R \Delta T}{n \cdot \frac{5}{2}R \Delta T} = \frac{3}{5}$$
So, percentage is $60\%$.
Fraction for $W$:
$$\frac{W}{Q} = \frac{nR \Delta T}{n \cdot \frac{5}{2}R \Delta T} = \frac{2}{5}$$
So, percentage is $40\%$.
Step 5: Ratio $\mathrm{A} : \mathrm{B}$
$\mathrm{A} : \mathrm{B} = 60\% : 40\% = 3 : 2$
Final Answer:
Option C: $3 : 2$

Answer

$5: 3$

Answer

$2: 3$

Answer

$3: 2$

Answer

$2: 5$

Thermodynamics & KTG

A monoatomic ideal gas is heated at constant pressure. The percentage of total heat used in increasing the internal energy and that used for doing external work is A and B respectively. Then the ratio, \mathrm{A}: \mathrm{B} is

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