Grade 12th passPhysical Chemistry

The PV-graph for a monoatomic gas is shown in the figure. Find the energy absorbed by the gas during the process.

Radhika

8 Years agoGrade 12th pass

7 Years ago

From first law of Thermodynamics,

$\Delta Q = \Delta U + \Delta W$

Lets find ∆W which is simply the area under the curve,

$\Delta W = \frac{1}{2}(2P_0 +3P_0)\times (2V_0 -V_0)$

$\Delta W = \frac{5}{2}P_0V_0$

Now lets find ∆U assuming one mole of gas, using,

$\Delta U = nC_V\Delta T =C_V \frac{(P_2V_2-P_1V_1)}{R}$

$\Delta U = \frac{3}{2}R \frac{(3P_02V_0-2P_0V_0)}{R} =6P_0V_0$

Note: Cv for monoatomic gas is 3R/2

So, $\Delta Q = 6P_0V_0+2.5P_0V_0=8.5P_0V_0$

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