**Problem (IIT JEE 2012):**
Two moles of ideal helium gas are in rubber balloon at $30\;\mathrm{{}^{o}C}$. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to $35\;\mathrm{{}^{o}C}$. The amount of heat required in raising the temperature is nearly, (Take $R=8.31\; \mathrm{J/mol-K}$).

- 62 J
- 104 J
- 124 J
- 208 J

**Solution:**
The pressure inside the balloon is equal to the constant ambient pressure because balloon is fully expandable. Thus, given process is isobaric, so heat exchanged in the process is given by,
\begin{align}
\label{mca:eqn:1}
\Delta Q=n C_p \Delta T.
\end{align}
The helium is a monoatomic gas with three degrees of freedom ($f=3$). Thus, the specific heat at constant volume ($C_v$) and constant pressure ($C_p$) for helium are given by,
\begin{align}
C_v&=({f}/{2})R=({3}/{2})R, \nonumber \\
C_p&=C_v+R=({3}/{2})R+R=({5}/{2})R. \nonumber
\end{align}
Substitute the values to get,
\begin{align}
\Delta Q=(2)(5/2)(8.31)(35-30)=207.75\approx 208\; \mathrm{J}. \nonumber
\end{align}

**Problem (IIT JEE 2009):**
$C_v$ and $C_p$ denotes the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then,

- $C_p-C_c$ is larger for a diatomic ideal gas than for a monoatomic ideal gas.
- $C_p+C_v$ is larger for a diatomic ideal gas than for a monoatomic ideal gas.
- $C_p/C_v$ is larger for a diatomic ideal gas than for a monoatomic ideal gas.
- $C_p\cdot C_v$ is larger for a diatomic ideal gas than for a monoatomic ideal gas.

**Solution:**
The molar heat capacities of an ideal gas having $f$ degrees of freedom are given by,
\begin{align}
\label{yea:eqn:1}
C_v&=R(f/2), \\
\label{yea:eqn:2}
C_p&=C_v+R=R{(f+2)}/{2}.
\end{align}
Use above equations to get,
\begin{align}
& C_p+C_v=(1+f)R,&& \frac{C_p}{C_v}=1+\frac{2}{f},&& C_p\cdot C_v=\frac{f(2+f)}{4}R^2.\nonumber
\end{align}
The value of $f$ for monoatomic and diatomic gases are 3 and 5. Thus, B and D are correct.