A cyclic process for 1 mole of an ideal gas is shown in figure in the V-T, diagram. The work done in A B, B C and C A respectively

Question

A cyclic process for 1 mole of an ideal gas is shown in figure in the V-T, diagram. The work done in A B, B C and C A respectively (A) 0, R T2 ln ((V1/V2)), R(T1-T2) (B) R(T1-T2), 0, R T1 ln (V1/V2) (C) 0, R T2 ln ((V2/V1)), R(T1-T2) (D) 0, R T2 ln ((V2/V1)), R(T2-T1)

Tardigrade Admin · Accepted Answer

Process A B is isochoric, ∴ WA B=P Δ V=0 Process B C is isothermal ∴ WB C=R T2 ⋅ ln ((V2/V1)) Process C A is isobaric ∴ WC A=-P Δ V=-R Δ T =-R(T1-T2)=R(T2-T1) (Negative sign is taken because of compression)

Q. A cyclic process for 1 mole of an ideal gas is shown in figure in the $V - T$ , diagram. The work done in $A B, BC$ and $C A$ respectively

$0, R T_{2} ln (\frac{V _{1}}{V _{2}}), R (T_{1} - T_{2})$

$R (T_{1} - T_{2}), 0, R T_{1} ln \frac{V _{1}}{V _{2}}$

$0, R T_{2} ln (\frac{V _{2}}{V _{1}}), R (T_{1} - T_{2})$

$0, R T_{2} ln (\frac{V _{2}}{V _{1}}), R (T_{2} - T_{1})$

Process $A B$ is isochoric,
$∴ W_{A B} = P Δ V = 0$
Process $BC$ is isothermal
$∴ W_{BC} = R T_{2} \cdot ln (\frac{V _{2}}{V _{1}})$
Process $C A$ is isobaric
$∴ W_{C A} = - P Δ V = - R Δ T$
$= - R (T_{1} - T_{2}) = R (T_{2} - T_{1})$
(Negative sign is taken because of compression)

Q. A cyclic process for 1 mole of an ideal gas is shown in figure in the V−T, diagram. The work done in AB,BC and CA respectively

0,RT2​ln(V2​V1​​),R(T1​−T2​)

R(T1​−T2​),0,RT1​lnV2​V1​​

0,RT2​ln(V1​V2​​),R(T1​−T2​)

0,RT2​ln(V1​V2​​),R(T2​−T1​)