Answer to Question #180742 in Molecular Physics | Thermodynamics for Tobias Amukwiita

Question

One mole of a monatomic perfect gas initially at temperature T expands from volume V to 2V, (a) at constant temperature, (b) at constant pressure. Calculate the work of expansion and the heat absorbed by the gas in each case? (5marks)

Answer 1

At constant Temperature

$dw = RT\ln{V_2\over V_1}$

$dw = RT\ln{2V\over V}$

$\boxed{dw = RT\ln{2}}$

from first law of thermodynamics

dQ = du + dw

since temperature is constant , change in internal energy du = 0

$\implies$

dQ= dw

$\therefore$ Heat absorbed $\boxed{dQ = RT\ln{2} }$

At constant pressure

The first law of thermodynamic says

$$\boxed{Q=\Delta U+W}$$

The change in internal energy of one mole of a monoatomic gas

$$\Delta U=\frac{3}{2}P(V_2-V_1)\\ =\frac{3}{2}P(2V_0-V_0)=\frac{3}{2}PV_0=\frac{3}{2}RT_0$$

The work done by a gas

$$\boxed{W=P(V_2-V_0)=PV_0=RT_0}$$

Finally

$$\boxed{Q=\frac{3}{2}RT_0+RT_0=\frac{5}{2}RT_0}$$