Question

Figure shows two rigid vessels A and B, each of volume 200 cm³, containing an ideal gas (C_v = 12.5 J K⁻¹ mol⁻¹). The vessels are connected to a manometer tube containing mercury. The pressure in both the vessels is 75 cm of mercury and the temperature is 300 K. (a) Find the number of moles of the gas in each vessel. (b) 5.0 J of heat is supplied to the gas in vessel A and 10 J to the gas in vessel B. Assuming there's no appreciable transfer of heat from A to B, calculate the difference in the heights of mercury in the two sides of the manometer. Gas constant, R = 8.3 J K⁻¹ mol⁻¹.

Answer 1

Given:
Volume of gas in each vessel, V = 200 cm³
Specific heat at constant volume of the gas, C_v = 12.5 J/mol-K
Initial temperature of the gas, T = 300 K
Initial pressure of the gas, P = 75 cm of Hg

(a) Using the ideal gas equation, number of moles of gases in each vessel,

$$\begin{gathered} 75\mathrm{cm}\mathrm{of}\mathrm{Hg}=99991.5\mathrm{N}/{\mathrm{m}}^2 \\ n\mathit{=}\frac{PV}{RT} \\ =\frac{99991.5\times 200\times {10}^{-6}}{8.3\times 300} \\ =8031.4\times {10}^{-6} \\ =0.008 \end{gathered}$$

(b) Heat is supplied to the gas, but dV is zero as the container has rigid walls.
So, dW = $P\mathit{∆}V$ = 0

From first law of thermodynamics,
dQ = dU
⇒ 5 = nC_vdT
⇒ 5 = 0.008 × 12.5 × dT
⇒ dT = 50 for A
$\frac{\mathit{P}}{\mathit{T}}=\frac{P_{\mathrm{A}}}{T_{\mathrm{A}}}$ because volume is kept constant.
$$\begin{gathered} Q=n{C}_{v}dT \\ T=\frac{Q}{n{C}_v} \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{75}{300}=\frac{{\mathrm{P}}_{\mathrm{A}}\times 0.008\times 12.5}{5} \\ \Rightarrow P_{\mathrm{A}}=\frac{75\times 5}{300\times 0.008\times 12.5} \\ =12.5\mathrm{cm}\mathrm{of}\mathrm{Hg} \\ \text{A}\mathrm{gain},\frac{P}{T}\mathit{=}\frac{P_B}{T_B}[\mathrm{For}\mathrm{container}\mathrm{B}] \\ \Rightarrow \frac{75}{300}=\frac{P_{\mathrm{B}}\times 0.008\times 12.5}{10} \end{gathered}$$
P_B = 25 cm of Hg

The distance moved by the mercury,
P_B − P_A = 25 − 12.5 = 12.5 cm