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Grade: 12th pass

                        

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

3 years ago

Answers : (1)

Samir Bhardwaj
34 Points
							
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
 
one year ago
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