- Home
- Standard 11
- Physics
$P-V$ diagram of $2\,g$ of $He$ gas for $A \to B$ process is shown. What is the heat given to the gas ?
$4\,P_0V_0$
$6\,P_0V_0$
$4.5\,P_0V_0$
$2\,P_0V_0$
Solution
$W=$ Area $=\frac{1}{2}\left(P_{0}+2 P_{0}\right) \cdot\left(2 V_{0}-V_{0}\right)=+\frac{3 P_{0} V_{0}}{2}$
$\mathrm{P}_{0} \mathrm{V}_{0}=\mu \mathrm{RT}_{1} \quad$ and $\left(2 \mathrm{P}_{0}\right)\left(2 \mathrm{V}_{0}\right)=\mu \mathrm{R} \mathrm{T}_{2}$
On subtracting $3 \mathrm{P}_{0} \mathrm{V}_{0}=\mu \mathrm{R} \Delta \mathrm{T}$
Now $\quad \Delta \mathrm{U}=\mu \mathrm{C}_{\mathrm{v}} \Delta \mathrm{T}=\mu\left(\frac{3}{2} \mathrm{R}\right) \times\left[\mathrm{T}_{2}-\mathrm{T}_{1}\right]$
$\Rightarrow \quad \frac{3}{2}(\mu R \Delta T)=\frac{9 P_{0} V_{0}}{2}$
$\mathrm{Q}=\mathrm{W}+\Delta \mathrm{U}=\frac{3 \mathrm{P}_{0} \mathrm{V}}{2}+\frac{9 \mathrm{P}_{0} \mathrm{V}_{0}}{2}=6 \mathrm{P}_{0} \mathrm{V}_{0}$
