Question

The $P-V$ diagram of $2\text{gm}$ of helium gas for a certain process is shown in the figure. Find the amount of heat given to the gas during the process ?

• A. $4P_0{V}_0$
• B. $6P_0{V}_0$
• C. $4.5P_0{V}_0$
• D. $2P_0{V}_0$

Answer 1

The correct option is B $6P_0{V}_0$

Since, Helium is a monoatomic gas,
Number of degrees of freedom $\left(f\right)=3$
To find the change in internal energy from A $\to$ B, we use
$\Delta U=\frac{f}{2}\mu R\Delta T$
where $\mu$ is the number of moles of helium gas.
Given, mass of helium $\left(m\right)=2\text{gm}$
Thus, the number of moles of helium gas
$\mu =\frac{\text{Mass}}{\text{Molecular weight}}=\frac{2}{2}=1$
Using ideal gas equation, we can write $\Delta \left(PV\right)=\mu R\Delta T$
i.e $\Delta U=\frac{f}{2}(P_f{V}_f-P_i{V}_i)$
Substituting the data given in the diagram,
$\Delta U=\frac{3}{2}(2P_0\times 2V_0-P_0\times V_0)=\frac{9}{2}{P}_0{V}_0$
Work done during process A $\to$ B is equal to the area enclosed by the graph above the volume axis i.e.
$W_{A\to B}=\frac{1}{2}(P_0+2P_0)\times (2V_0-V_0)=\frac{3}{2}{P}_0{V}_0$
Hence, using first law of thermodynamics,
$\Delta Q=\Delta U+\Delta W=\frac{9}{2}{P}_0{V}_0+\frac{3}{2}{P}_0{V}_0=6P_0{V}_0$
Hence, option (b) is the correct answer.