Question

At constant pressure, equal amounts of heat are supplied to a monatomic gas and a diatomic gas separately. The ratio of the increases in internal energies of the two gases is

• A. 1:1
• B. 9:49
• C. 3:7
• D. 21:25

Hint:

The change in internal energy $\Delta U$ always depends on $C_v$ ($\Delta U = nC_v\Delta T$), regardless of the process. The heat supplied $Q$ depends on the process ($Q=nC_p\Delta T$ for isobaric). The ratio $\Delta U / Q_p = C_v/C_p = 1/\gamma$.

Answer 1

The Correct Option is D

Let the heat supplied at constant pressure be $Q_p$.
$Q_p = nC_p \Delta T$. Since equal heat is supplied to the same number of moles (implied by context), we have $C_{p1}\Delta T_1 = C_{p2}\Delta T_2$.
The increase in internal energy is given by $\Delta U = nC_v \Delta T$.
We need to find the ratio $\frac{\Delta U_1}{\Delta U_2} = \frac{nC_{v1}\Delta T_1}{nC_{v2}\Delta T_2} = \frac{C_{v1}\Delta T_1}{C_{v2}\Delta T_2}$.
From $Q_p$ being equal, we have $\frac{\Delta T_1}{\Delta T_2} = \frac{C_{p2}}{C_{p1}}$.
Substitute this into the ratio for $\Delta U$: $\frac{\Delta U_1}{\Delta U_2} = \frac{C_{v1}}{C_{v2}} \cdot \frac{C_{p2}}{C_{p1}} = \left(\frac{C_{v1}}{C_{p1}}\right) \cdot \left(\frac{C_{p2}}{C_{v2}}\right)$.
The ratio $\frac{C_p}{C_v}$ is the adiabatic index, $\gamma$. So, $\frac{C_v}{C_p} = \frac{1}{\gamma}$.
$\frac{\Delta U_1}{\Delta U_2} = \frac{1/\gamma_1}{1/\gamma_2} = \frac{\gamma_2}{\gamma_1}$. This logic is flawed. Let's restart.
Let $Q$ be the heat supplied. $Q = n C_p \Delta T$. The increase in internal energy is $\Delta U = n C_v \Delta T$.
From the first equation, $\Delta T = \frac{Q}{nC_p}$.
Substitute this into the second equation: $\Delta U = n C_v \left(\frac{Q}{nC_p}\right) = Q \frac{C_v}{C_p} = \frac{Q}{\gamma}$.
So, for a given amount of heat $Q$ supplied at constant pressure, the change in internal energy is $\Delta U = Q/\gamma$.
The ratio of the increases in internal energies is:
$\frac{\Delta U_{mono}}{\Delta U_{di}} = \frac{Q/\gamma_{mono}}{Q/\gamma_{di}} = \frac{\gamma_{di}}{\gamma_{mono}}$.
For a monatomic gas (Gas 1), degrees of freedom $f=3$. $\gamma_{mono} = \frac{f+2}{f} = \frac{3+2}{3} = \frac{5}{3}$.
For a diatomic gas (Gas 2), degrees of freedom $f=5$ (at ordinary temperatures). $\gamma_{di} = \frac{f+2}{f} = \frac{5+2}{5} = \frac{7}{5}$.
$\frac{\Delta U_{mono}}{\Delta U_{di}} = \frac{7/5}{5/3} = \frac{7}{5} \times \frac{3}{5} = \frac{21}{25}$.
The ratio is 21:25.