Question

An ideal gas is in equilibrium at initial state with temperature
T=137oCT=137oC
, pressure P = 0.75Pa and volume
V=0.75m3V=0.75m3
. If there is a change in state in which the gas undergoes an isothermal process to a final state of equilibrium during which the volume is doubled. Calculate the temperature and pressure of the gas at this final state.

187oC187oC
and
0.19m30.19m3

394oC394oC
and
0.38m30.38m3

187oC187oC
and
0.38m30.38m3

394oC394oC
and
0.19m30.19m3

Answer 1

Answer on Question # 58712 – Physics – Molecular Physics | Thermodynamics

An ideal gas is in equilibrium at initial state with temperature $T_0 = 137 \, {}^\circ \text{C}$, pressure $P_0 = 0.75 \, \text{Pa}$ and volume $V_0 = 0.75 \, \text{m}^3$. If there is a change in state in which the gas undergoes an isothermal process to a final state of equilibrium during which the volume is doubled. Calculate the temperature and pressure of the gas at this final state.

Solution:

In an isothermal process, the temperature of the gas is constant. The temperature of the gas at the final state $T_f$ is equal to the initial temperature:

$$T_f = T_0 = 137 \, \left[ {}{}^\circ \text{C} \right].$$

The relation between thermodynamic parameters in an isothermal process is described by Boyle-Mariotte law:

$$pV = \text{const};$$ $$p_0 V_0 = p_f V_f.$$

The pressure of the gas in the final state:

$$p_f = \frac{p_0 V_0}{V_f} = \frac{p_0 V_0}{2 V_0} = \frac{p_0}{2} = \frac{0.75}{2} = 0.375 \, \left[ \text{Pa} \right].$$

Answer: $T_f = 137 \, \left[ {}{}^\circ \text{C} \right]$; $p_f = 0.375 \, \left[ \text{Pa} \right]$.

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