When 80 J of heat is absorbed by a monotonic gas, its volume increases by $16 \times 10^{-5} \mathrm{~m}^3$.…

When 80 J of heat is absorbed by a monotonic gas, its volume increases by $16 \times 10^{-5} \mathrm{~m}^3$. The pressure of the gas is
  1. $2 \times 10^5 \mathrm{Nm}^{-2}$
  2. $4 \times 10^5 \mathrm{Nm}^{-2}$
  3. $6 \times 10^5 \mathrm{Nm}^{-2}$
  4. $5 \times 10^5 \mathrm{Nm}^{-2}$

Solution

For monoatomic gas, $\mathrm{f}=3$ $\therefore \Delta \mathrm{u}=\frac{\mathrm{nfR} \Delta \mathrm{~T}}{2}=\frac{3}{2} \mathrm{n} \mathrm{R} \Delta \mathrm{~T}$
By first law of thermodynamics, $\begin{aligned} & \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W}=-\mathrm{nR} \Delta \mathrm{~T}+\mathrm{nR} \Delta \mathrm{~T}=\frac{5}{2} \mathrm{nR} \Delta \mathrm{~T} \\ & \Rightarrow 80=\frac{5}{2} \mathrm{nR} \Delta \mathrm{~T}=\frac{5}{2} \mathrm{p} \Delta \mathrm{v}=\frac{5}{2} \times 16 \times 10^{-5} \times \mathrm{p} \\ & \therefore \mathrm{P}=2 \times 10^5 \mathrm{Nm}^{-2} \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 2)

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