Answer to Question #131436 in Molecular Physics | Thermodynamics for kshitiz

Question

One mole of a monoatomic perfect gas is initially at a constant temperature T0 . It expands from a volume V0 to 2V0 under constant pressure. The heat absorbed by the gas is

Answer 1

The first law of thermodynamic says

$$Q=\Delta U+W$$

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

$$W=P(V_2-V_0)=PV_0=RT_0$$

Finally

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