First slide

Thermodynamic processes

Question

Three moles of an ideal gas are taken through a cyclic process ABCA as shown on T-V diagram in Fig. The gas loses 2510 J of heat in the complete cycle. If $T_{A} = 100 K and T_{B} = 200 K$ . The work done by the gas during the process BC is….. (Take R = 8.3 JmolK^-1)

Moderate

A

5000J
B

-5000J
C

2490J
D

-2510J

Solution

In the process AB, the volume V increases linearly with temperature T. hence AB is isobaric (constant pressure). Therefore, work done in this process is
$W_{AB} = P ∆ V = nR ∆ T (∵ PV = nRT)$
$= nR (T_{B} - T_{A})$
$= 3 \times 8.3 \times (200 - 100) = 2490 J$
Process CA is isochoric (constant volume). Hence work done in this process W_CA = 0. Since the whole process ABCA is cyclic, the change in internal energy in the complete cycle is zero, i.e., $∆ U = 0$ . Now, from the first law of the thermodynamics, (Given Q = -2510 J)
$Q = ∆ U + W = ∆ U + W_{AB} + W_{BC} + W_{CA}$
$- 2510 = 0 + 2490 + W_{BC} + 0$
$W_{BC} = - 2510 - 2490 = - 5000 J$
The negative sign shows that the work is done by the gas.

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