(A) For an ideal gas,the equation of state is $PV = nRT$,which can be written as $P = nRT(1/V)$.Along $AB$,the pressure $P$ is constant,and the value

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Along $AB$,temperature decreases while along $BC$ temperature increases.

(A) For an ideal gas,the equation of state is $PV = nRT$,which can be written as $P = nRT(1/V)$.Along $AB$,the pressure $P$ is constant,and the value of $(1/V)$ increases. Since $P = nRT(1/V)$,if $P$ is constant and $(1/V)$ increases,the temperature $T$ must decrease.Along $BC$,the value of $(1/V)$ is constant,and the pressure $P$ increases. Since $P = nRT(1/V)$,if $(1/V)$ is constant and $P$ increases,the temperature $T$ must increase.Therefore,the correct statement is that along $AB$,temperature decreases while along $BC$ temperature increases.

Class 11 Physics Thermodynamics Heat, Work done and Internal Energy from Graph

Difficult

An enclosed ideal gas is taken through a cycle as shown in the figure. Then

A
Along $AB$,temperature decreases while along $BC$ temperature increases.
B
Along $AB$,temperature increases while along $BC$ the temperature decreases.
C
Along $CA$ work is done by the gas and the internal energy remains constant.
D
Along $CA$ work is done on the gas and internal energy of the gas increases.

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Thermodynamics

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Heat, Work done and Internal Energy from Graph

For one complete cycle of a thermodynamic process on a gas as shown in the $P-V$ diagram,which of the following is correct?

Easy

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$A$ system goes from $A$ to $B$ via two processes $I$ and $II$ as shown in the figure. If $\Delta U_1$ and $\Delta U_2$ are the changes in internal energies in the processes $I$ and $II$ respectively,then

Medium

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$A$ thermodynamic system is taken from an original state $A$ to an intermediate state $B$ by a linear process as shown in the figure. Its volume is then reduced to the original value from $B$ to $C$ by an isobaric process. The total work done by the gas from $A$ to $B$ and $B$ to $C$ is: (in $J$)

DifficultJEE MAIN 2024

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In the given $PV$ diagram,a system goes from state $P$ to state $Q$ along two different paths. Along path $1$,the system is given $1000 \ J$ of heat. The work done by the system along path $1$ is $100 \ J$ more than the work done along path $2$. How much heat is transferred along path $2$ (in $J$)?

Difficult

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$A$ sample of an ideal gas is taken through the cyclic process $abca$ as shown in the figure. The change in the internal energy of the gas along the path $ca$ is $-180\, J$. The gas absorbs $250\, J$ of heat along the path $ab$ and $60\, J$ along the path $bc$. The work done by the gas along the path $abc$ is ..... $J$.

MediumJEE MAIN 2019

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An enclosed ideal gas is taken through a cycle as shown in the figure. Then

Similar Questions

For one complete cycle of a thermodynamic process on a gas as shown in the $P-V$ diagram,which of the following is correct?

$A$ system goes from $A$ to $B$ via two processes $I$ and $II$ as shown in the figure. If $\Delta U_1$ and $\Delta U_2$ are the changes in internal energies in the processes $I$ and $II$ respectively,then

$A$ thermodynamic system is taken from an original state $A$ to an intermediate state $B$ by a linear process as shown in the figure. Its volume is then reduced to the original value from $B$ to $C$ by an isobaric process. The total work done by the gas from $A$ to $B$ and $B$ to $C$ is: (in $J$)

In the given $PV$ diagram,a system goes from state $P$ to state $Q$ along two different paths. Along path $1$,the system is given $1000 \ J$ of heat. The work done by the system along path $1$ is $100 \ J$ more than the work done along path $2$. How much heat is transferred along path $2$ (in $J$)?

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