Question

N moles of an ideal gas undergoes adiabatic process and changes its state from P1,V1,T1 to P2,V2,T2 . The change in internal energy of the gas will be​

Answer 1

The ideal gas undergoes adiabatic process, so no heat enters or leaves the system, i.e., $\tt{\Delta q=0.}$

By first law of thermodynamics,

$\tt{\longrightarrow \Delta u=\Delta q+\Delta W}$

or,

$\tt{\longrightarrow \Delta u=\Delta W\quad\quad[\because\,\Delta q=0]}$

I.e., the work done on the system is completely utilized by the system to change its internal energy.

Work done in an adiabatic process is,

$\tt{\longrightarrow \Delta W=\dfrac{P_1V_1-P_2V_2}{1-\gamma}}$

This is the same as the internal energy.

$\tt{\longrightarrow\underline{\underline{\Delta u=\dfrac{P_1V_1-P_2V_2}{1-\gamma}}}}$

In terms of temperature,

$\tt{\longrightarrow\underline{\underline{\Delta u=\dfrac{NR(T_1-T_2)}{1-\gamma}}}}$

Answer 2

ANSWER IS -------------------->>>>>>>>>>>>

The ideal gas undergoes adiabatic process, so no heat enters or leaves the system, i.e., \tt{\Delta q=0.}Δq=0.

The ideal gas undergoes adiabatic process, so no heat enters or leaves the system, i.e., \tt{\Delta q=0.}Δq=0.By first law of thermodynamics,

The ideal gas undergoes adiabatic process, so no heat enters or leaves the system, i.e., \tt{\Delta q=0.}Δq=0.By first law of thermodynamics,\tt{\longrightarrow \Delta u=\Delta q+\Delta W}⟶Δu=Δq+ΔW

The ideal gas undergoes adiabatic process, so no heat enters or leaves the system, i.e., \tt{\Delta q=0.}Δq=0.By first law of thermodynamics,\tt{\longrightarrow \Delta u=\Delta q+\Delta W}⟶Δu=Δq+ΔWor,

The ideal gas undergoes adiabatic process, so no heat enters or leaves the system, i.e., \tt{\Delta q=0.}Δq=0.By first law of thermodynamics,\tt{\longrightarrow \Delta u=\Delta q+\Delta W}⟶Δu=Δq+ΔWor,\tt{\longrightarrow \Delta u=\Delta W\quad\quad[\because\,\Delta q=0]}⟶Δu=ΔW[∵Δq=0]

The ideal gas undergoes adiabatic process, so no heat enters or leaves the system, i.e., \tt{\Delta q=0.}Δq=0.By first law of thermodynamics,\tt{\longrightarrow \Delta u=\Delta q+\Delta W}⟶Δu=Δq+ΔWor,\tt{\longrightarrow \Delta u=\Delta W\quad\quad[\because\,\Delta q=0]}⟶Δu=ΔW[∵Δq=0]I.e., the work done on the system is completely utilized by the system to change its internal energy.

The ideal gas undergoes adiabatic process, so no heat enters or leaves the system, i.e., \tt{\Delta q=0.}Δq=0.By first law of thermodynamics,\tt{\longrightarrow \Delta u=\Delta q+\Delta W}⟶Δu=Δq+ΔWor,\tt{\longrightarrow \Delta u=\Delta W\quad\quad[\because\,\Delta q=0]}⟶Δu=ΔW[∵Δq=0]I.e., the work done on the system is completely utilized by the system to change its internal energy.Work done in an adiabatic process is,

The ideal gas undergoes adiabatic process, so no heat enters or leaves the system, i.e., \tt{\Delta q=0.}Δq=0.By first law of thermodynamics,\tt{\longrightarrow \Delta u=\Delta q+\Delta W}⟶Δu=Δq+ΔWor,\tt{\longrightarrow \Delta u=\Delta W\quad\quad[\because\,\Delta q=0]}⟶Δu=ΔW[∵Δq=0]I.e., the work done on the system is completely utilized by the system to change its internal energy.Work done in an adiabatic process is,\tt{\longrightarrow \Delta W=\dfrac{P_1V_1-P_2V_2}{1-\gamma}}⟶ΔW=1−γP1V1−P2V2

The ideal gas undergoes adiabatic process, so no heat enters or leaves the system, i.e., \tt{\Delta q=0.}Δq=0.By first law of thermodynamics,\tt{\longrightarrow \Delta u=\Delta q+\Delta W}⟶Δu=Δq+ΔWor,\tt{\longrightarrow \Delta u=\Delta W\quad\quad[\because\,\Delta q=0]}⟶Δu=ΔW[∵Δq=0]I.e., the work done on the system is completely utilized by the system to change its internal energy.Work done in an adiabatic process is,\tt{\longrightarrow \Delta W=\dfrac{P_1V_1-P_2V_2}{1-\gamma}}⟶ΔW=1−γP1V1−P2V2This is the same as the internal energy.

The ideal gas undergoes adiabatic process, so no heat enters or leaves the system, i.e., \tt{\Delta q=0.}Δq=0.By first law of thermodynamics,\tt{\longrightarrow \Delta u=\Delta q+\Delta W}⟶Δu=Δq+ΔWor,\tt{\longrightarrow \Delta u=\Delta W\quad\quad[\because\,\Delta q=0]}⟶Δu=ΔW[∵Δq=0]I.e., the work done on the system is completely utilized by the system to change its internal energy.Work done in an adiabatic process is,\tt{\longrightarrow \Delta W=\dfrac{P_1V_1-P_2V_2}{1-\gamma}}⟶ΔW=1−γP1V1−P2V2This is the same as the internal energy.\tt{\longrightarrow\underline{\underline{\Delta u=\dfrac{P_1V_1-P_2V_2}{1-\gamma}}}}⟶Δu=1−γP1V1−P2V2

The ideal gas undergoes adiabatic process, so no heat enters or leaves the system, i.e., \tt{\Delta q=0.}Δq=0.By first law of thermodynamics,\tt{\longrightarrow \Delta u=\Delta q+\Delta W}⟶Δu=Δq+ΔWor,\tt{\longrightarrow \Delta u=\Delta W\quad\quad[\because\,\Delta q=0]}⟶Δu=ΔW[∵Δq=0]I.e., the work done on the system is completely utilized by the system to change its internal energy.Work done in an adiabatic process is,\tt{\longrightarrow \Delta W=\dfrac{P_1V_1-P_2V_2}{1-\gamma}}⟶ΔW=1−γP1V1−P2V2This is the same as the internal energy.\tt{\longrightarrow\underline{\underline{\Delta u=\dfrac{P_1V_1-P_2V_2}{1-\gamma}}}}⟶Δu=1−γP1V1−P2V2In terms of temperature,

The ideal gas undergoes adiabatic process, so no heat enters or leaves the system, i.e., \tt{\Delta q=0.}Δq=0.By first law of thermodynamics,\tt{\longrightarrow \Delta u=\Delta q+\Delta W}⟶Δu=Δq+ΔWor,\tt{\longrightarrow \Delta u=\Delta W\quad\quad[\because\,\Delta q=0]}⟶Δu=ΔW[∵Δq=0]I.e., the work done on the system is completely utilized by the system to change its internal energy.Work done in an adiabatic process is,\tt{\longrightarrow \Delta W=\dfrac{P_1V_1-P_2V_2}{1-\gamma}}⟶ΔW=1−γP1V1−P2V2This is the same as the internal energy.\tt{\longrightarrow\underline{\underline{\Delta u=\dfrac{P_1V_1-P_2V_2}{1-\gamma}}}}⟶Δu=1−γP1V1−P2V2In terms of temperature,\tt{\longrightarrow\underline{\underline{\Delta u=\dfrac{NR(T_1-T_2)}{1-\gamma}}}}⟶Δu=1−γNR(T1−T2)

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