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Question
One mole of an ideal monoatomic gas undergoes a linear process from A to B, in which its pressure P and volume V changes as shown in figure
Choose the correct options.
One mole of an ideal monoatomic gas undergoes a linear process from A to B, in which its pressure P and volume V changes as shown in figure
Choose the correct options.
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Solution
The correct options are
A The maximum temperature of the gas during the process is PoVo4R
D As the volume of the gas is increased, then the volume after which (If the volume of the gas is further increased) the given process switches from endothermic to exothermic is 5Vo8
From the P−V graph, the relation between P and V is
P=−PoVoV+Po(i)
Using ideal gas equation for 1 mole,
PV=RT
Equation (i) becomes
T=PoRV[1−VVo](ii)
Differentiating above equation w.r.t V and equating it to zero to get the volume at which maximum temperature will be obtained,
dTdV=PoR[1−2VVo]=0
⇒V=Vo2
Using this value in equation (ii) to get maximum temperature
T=PoVo4R
As we know, Q=ΔU+W
where, ΔU is the change in internal energy of the gas and W is work done by the gas.
For one mole of monoatomic ideal gas, ΔU=32RΔT
The work done by the gas can be calculated from the area under the curve in the P−V graph.
Therefore, for the process AC as indicated in the graph below, the heat transfer can be calculated.
Let the temperature of the gas at state C be T.
Internal energy change for process AC, ΔU=32R(T−T1)=32(PV−P1V1)
Work done (W) by the gas for process AC will be the area under the line AC on P−V diagram
W=12(P+P1)(V−V1)
Thus, heat transfer, Q=ΔU+W
⇒Q=32(PV−P1V1)+12(P+P1)(V−V1)
Q=2PV−2P1V1−12[PV1−P1V]
Using equation (i)
i.e P=−PoVoV+Po
Q=2Po[1−VVo]V−2Po[1−V1Vo]V1−12Po[1−VVo]V1+12Po[1−V1Vo]V
On simplification we get,
Q=−2PoV2Vo+52PoV−2PoV1[54−V1Vo]
Now the process changes from endothermic to exothermic when dQdV changes from positive to negative i.e
at dQdV=0
Thus, dQdV=−4PoVVo+52Po=0
⇒V=58Vo
Hence the process changes from endothermic to exothermic when volume of gas reaches 58Vo
A The maximum temperature of the gas during the process is PoVo4R
D As the volume of the gas is increased, then the volume after which (If the volume of the gas is further increased) the given process switches from endothermic to exothermic is 5Vo8
From the P−V graph, the relation between P and V is
P=−PoVoV+Po(i)
Using ideal gas equation for 1 mole,
PV=RT
Equation (i) becomes
T=PoRV[1−VVo](ii)
Differentiating above equation w.r.t V and equating it to zero to get the volume at which maximum temperature will be obtained,
dTdV=PoR[1−2VVo]=0
⇒V=Vo2
Using this value in equation (ii) to get maximum temperature
T=PoVo4R
As we know, Q=ΔU+W
where, ΔU is the change in internal energy of the gas and W is work done by the gas.
For one mole of monoatomic ideal gas, ΔU=32RΔT
The work done by the gas can be calculated from the area under the curve in the P−V graph.
Therefore, for the process AC as indicated in the graph below, the heat transfer can be calculated.
Let the temperature of the gas at state C be T.
Internal energy change for process AC, ΔU=32R(T−T1)=32(PV−P1V1)
Work done (W) by the gas for process AC will be the area under the line AC on P−V diagram
W=12(P+P1)(V−V1)
Thus, heat transfer, Q=ΔU+W
⇒Q=32(PV−P1V1)+12(P+P1)(V−V1)
Q=2PV−2P1V1−12[PV1−P1V]
Using equation (i)
i.e P=−PoVoV+Po
Q=2Po[1−VVo]V−2Po[1−V1Vo]V1−12Po[1−VVo]V1+12Po[1−V1Vo]V
On simplification we get,
Q=−2PoV2Vo+52PoV−2PoV1[54−V1Vo]
Now the process changes from endothermic to exothermic when dQdV changes from positive to negative i.e
at dQdV=0
Thus, dQdV=−4PoVVo+52Po=0
⇒V=58Vo
Hence the process changes from endothermic to exothermic when volume of gas reaches 58Vo
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