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Question: Carbon monoxide is carried around a closed cycle abc in which bc is an isothermal process as shown i...

Carbon monoxide is carried around a closed cycle abc in which bc is an isothermal process as shown in the figure. The gas absorbs 7000  J7000\;{\rm{J}} of heat as its temperature increases from 300  K300\;{\rm{K}} to 1000  K1000\;{\rm{K}} in going from a to b. the quantity of heat rejected by the gas during the process ca is
(a). 4200  J4200\;{\rm{J}}
(b). 5000  J5000\;{\rm{J}}
(c). 9000  J9000\;{\rm{J}}
(d). 9800  J9800\;{\rm{J}}

Explanation

Solution

We should remember that if there is an isobaric process, the pressure should not vary. Similarly, we should also keep in mind that volume of a system will be constant in an isobaric process.

Complete step by step answer:
As the volume remains constant from a to b, the process from a to b is isochoric. For a system the change in heat energy is dependent on the internal energy and the total work done. Thus for isochoric process from a to b, we have,
ΔQab=ΔUab+Δwab\Delta {Q_{ab}} = \Delta {U_{ab}} + \Delta {w_{ab}}
Here, UU is the internal energy and ww is the work done. But for an isochoric process, as the volume remains the same, the total change in work done is zero. That is, Δw=0\Delta w = 0. Thus , from a to b,
ΔQab=ΔUab\Delta {Q_{ab}} = \Delta {U_{ab}}
It is given that the gas absorbs 7000  J7000\;{\rm{J}} of heat, which is ΔQab\Delta {Q_{ab}}, we have,
ΔQab=7000  J=ΔUab\Delta {Q_{ab}} = 7000\;{\rm{J}} = \Delta {U_{ab}}
Now the formula to calculate the internal energy change is,
ΔU=nCvΔT\Delta U = n{C_v}\Delta T
Here Cv{C_v} is the molar heat capacity at constant volume, nn is the number of moles and TT is the temperature of the system.
Here as the gas considered is carbon monoxide, which is a diatomic gas, we have,
Cv=52R{C_v} = \dfrac{5}{2}R
Here RR is the gas constant.
Also it is given that the temperature increases from 300  K300\;{\rm{K}} to 1000  K1000\;{\rm{K}}. Thus the difference in temperature is 1000  K300  K=700  K1000\;{\rm{K}} - 300\;{\rm{K = 700}}\;{\rm{K}}. Thus by substituting for Cv{C_v},and temperature,
ΔU=n52RΔT 7000  J=nR700×52 nR=4 \Delta U = n\dfrac{5}{2}R\Delta T\\\ \Rightarrow 7000\;{\rm{J}} = nR\dfrac{{700 \times 5}}{2}\\\ \Rightarrow nR = 4
Let us now consider the change from the point c to a. here the pressure is remaining constant. So the process is isobaric. Thus the change in heat energy from the point c to a is calculated as,
ΔQ=nCpΔT\Delta Q = n{C_p}\Delta T
Here Cp{C_p} is the molar heat capacity at constant pressure, nn is the number of moles and TT is the temperature of the system.
As the gas considered is carbon monoxide, which is a diatomic gas, we have,
Cp=72R{C_p} = \dfrac{7}{2}R
Also it is given that the bc is an isothermal process. So the temperature for b will be same as that of temperature for c. So the change in temperature for the isobaric process ca is 300  K1000  K=700  K300\;{\rm{K}} - {\rm{1000}}\;{\rm{K}} = {\rm{700}}\;{\rm{K}}. Thus substituting for Cp{C_p} and the temperature difference in the equation ΔQ=nCpΔT\Delta Q = n{C_p}\Delta T, we have,
ΔQ=n72R700  K ΔQ=nR72×700  K \Delta Q = - n\dfrac{7}{2}R700\;{\rm{K}}\\\ \Rightarrow\Delta Q {\rm{ = - nR}}\dfrac{7}{2} \times 700\;{\rm{K}}
Substituting 44 for nRnR, we have,
ΔQ=4×72×700  K ΔQ=9800  J \Delta Q = - 4 \times \dfrac{7}{2} \times 700\;{\rm{K}}\\\ \therefore\Delta Q = - 9800\;{\rm{J}}
Therefore the heat rejected is 9800  J9800\;{\rm{J}} and the correct option is(d).

Note: The molar heat capacity of a gas at constant volume is defined as the heat energy required to warm one mole of a gas through one degree when its volume is kept constant. And at constant volume we get the molar heat capacity Cp{C_p}. And the gas constant is defined as the difference between the molar heat capacities at constant pressure and volume.