272728
1 mole of gas occupying 3L volume is expanded against a constant external pressure of 1 atm to a volume of $15 \mathrm{~L}$. The work done by a system is equal to
272729
When 1 mole of a gas is heated at constant volume, temperature is raised form $298 \mathrm{~K}$ to 308K. Heat supplied to the gas is $500 \mathrm{~J}$. Then, which statement is correct?
We know that, $\Delta \mathrm{H}=\Delta \mathrm{E}+\mathrm{P} \Delta \mathrm{V}$ From the first law of thermodynamics, $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ $\mathrm{W}=-\mathrm{P} \Delta \mathrm{V}$ $\Delta \mathrm{V}=0\{\text { Given in question }\}$ $\mathrm{W}=-\mathrm{P} \times 0=0$ $\mathrm{q}=500 \mathrm{~J}$ So, $\quad \Delta \mathrm{U}=\mathrm{q}=500 \mathrm{~J}$
UPTU/UPSEE-2014
Thermodynamics
272730
Pressure-volume (PV) work done by an ideal gaseous system at constant volume is (where $E$ is internal energy of the system)
1 $-\Delta \mathrm{P} / \mathrm{P}$
2 zero
3 $-\mathrm{V} \Delta \mathrm{P}$
4 $-\Delta \mathrm{E}$
Explanation:
From the first law of thermodynamics $\Delta \mathrm{E}=\mathrm{q}+\mathrm{W}$ $\mathrm{W}=-\mathrm{P} \Delta \mathrm{V}$ $\therefore \Delta \mathrm{V}=0$ since the volume is constant $\mathrm{W}=0$
WB-JEE-2013
Thermodynamics
272660
Five moles of a gas are put through a series of changes as shown graphically in a cyclic process. The processes $A \rightarrow B, B \rightarrow C$ and $C \rightarrow$ A respectively are
1 isochoric, isobaric, isothermal
2 isobaric, isochoric, isothermal
3 isothermal, isobaric, isochoric
4 isochoric, isothermal, isobaric
Explanation:
Process $\mathrm{A} \rightarrow \mathrm{B}$ is isochoric, i.e., volume remains constant.
Process $\mathrm{B} \rightarrow \mathrm{C}$ is isobaric, i.e., pressure remains constant. ( $\because$ at constant $P, V \propto T$ )
Process $\mathrm{C} \rightarrow \mathrm{A}$ is isothermal, i.e., temperature remains constant.
272728
1 mole of gas occupying 3L volume is expanded against a constant external pressure of 1 atm to a volume of $15 \mathrm{~L}$. The work done by a system is equal to
272729
When 1 mole of a gas is heated at constant volume, temperature is raised form $298 \mathrm{~K}$ to 308K. Heat supplied to the gas is $500 \mathrm{~J}$. Then, which statement is correct?
We know that, $\Delta \mathrm{H}=\Delta \mathrm{E}+\mathrm{P} \Delta \mathrm{V}$ From the first law of thermodynamics, $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ $\mathrm{W}=-\mathrm{P} \Delta \mathrm{V}$ $\Delta \mathrm{V}=0\{\text { Given in question }\}$ $\mathrm{W}=-\mathrm{P} \times 0=0$ $\mathrm{q}=500 \mathrm{~J}$ So, $\quad \Delta \mathrm{U}=\mathrm{q}=500 \mathrm{~J}$
UPTU/UPSEE-2014
Thermodynamics
272730
Pressure-volume (PV) work done by an ideal gaseous system at constant volume is (where $E$ is internal energy of the system)
1 $-\Delta \mathrm{P} / \mathrm{P}$
2 zero
3 $-\mathrm{V} \Delta \mathrm{P}$
4 $-\Delta \mathrm{E}$
Explanation:
From the first law of thermodynamics $\Delta \mathrm{E}=\mathrm{q}+\mathrm{W}$ $\mathrm{W}=-\mathrm{P} \Delta \mathrm{V}$ $\therefore \Delta \mathrm{V}=0$ since the volume is constant $\mathrm{W}=0$
WB-JEE-2013
Thermodynamics
272660
Five moles of a gas are put through a series of changes as shown graphically in a cyclic process. The processes $A \rightarrow B, B \rightarrow C$ and $C \rightarrow$ A respectively are
1 isochoric, isobaric, isothermal
2 isobaric, isochoric, isothermal
3 isothermal, isobaric, isochoric
4 isochoric, isothermal, isobaric
Explanation:
Process $\mathrm{A} \rightarrow \mathrm{B}$ is isochoric, i.e., volume remains constant.
Process $\mathrm{B} \rightarrow \mathrm{C}$ is isobaric, i.e., pressure remains constant. ( $\because$ at constant $P, V \propto T$ )
Process $\mathrm{C} \rightarrow \mathrm{A}$ is isothermal, i.e., temperature remains constant.
272728
1 mole of gas occupying 3L volume is expanded against a constant external pressure of 1 atm to a volume of $15 \mathrm{~L}$. The work done by a system is equal to
272729
When 1 mole of a gas is heated at constant volume, temperature is raised form $298 \mathrm{~K}$ to 308K. Heat supplied to the gas is $500 \mathrm{~J}$. Then, which statement is correct?
We know that, $\Delta \mathrm{H}=\Delta \mathrm{E}+\mathrm{P} \Delta \mathrm{V}$ From the first law of thermodynamics, $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ $\mathrm{W}=-\mathrm{P} \Delta \mathrm{V}$ $\Delta \mathrm{V}=0\{\text { Given in question }\}$ $\mathrm{W}=-\mathrm{P} \times 0=0$ $\mathrm{q}=500 \mathrm{~J}$ So, $\quad \Delta \mathrm{U}=\mathrm{q}=500 \mathrm{~J}$
UPTU/UPSEE-2014
Thermodynamics
272730
Pressure-volume (PV) work done by an ideal gaseous system at constant volume is (where $E$ is internal energy of the system)
1 $-\Delta \mathrm{P} / \mathrm{P}$
2 zero
3 $-\mathrm{V} \Delta \mathrm{P}$
4 $-\Delta \mathrm{E}$
Explanation:
From the first law of thermodynamics $\Delta \mathrm{E}=\mathrm{q}+\mathrm{W}$ $\mathrm{W}=-\mathrm{P} \Delta \mathrm{V}$ $\therefore \Delta \mathrm{V}=0$ since the volume is constant $\mathrm{W}=0$
WB-JEE-2013
Thermodynamics
272660
Five moles of a gas are put through a series of changes as shown graphically in a cyclic process. The processes $A \rightarrow B, B \rightarrow C$ and $C \rightarrow$ A respectively are
1 isochoric, isobaric, isothermal
2 isobaric, isochoric, isothermal
3 isothermal, isobaric, isochoric
4 isochoric, isothermal, isobaric
Explanation:
Process $\mathrm{A} \rightarrow \mathrm{B}$ is isochoric, i.e., volume remains constant.
Process $\mathrm{B} \rightarrow \mathrm{C}$ is isobaric, i.e., pressure remains constant. ( $\because$ at constant $P, V \propto T$ )
Process $\mathrm{C} \rightarrow \mathrm{A}$ is isothermal, i.e., temperature remains constant.
272728
1 mole of gas occupying 3L volume is expanded against a constant external pressure of 1 atm to a volume of $15 \mathrm{~L}$. The work done by a system is equal to
272729
When 1 mole of a gas is heated at constant volume, temperature is raised form $298 \mathrm{~K}$ to 308K. Heat supplied to the gas is $500 \mathrm{~J}$. Then, which statement is correct?
We know that, $\Delta \mathrm{H}=\Delta \mathrm{E}+\mathrm{P} \Delta \mathrm{V}$ From the first law of thermodynamics, $\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}$ $\mathrm{W}=-\mathrm{P} \Delta \mathrm{V}$ $\Delta \mathrm{V}=0\{\text { Given in question }\}$ $\mathrm{W}=-\mathrm{P} \times 0=0$ $\mathrm{q}=500 \mathrm{~J}$ So, $\quad \Delta \mathrm{U}=\mathrm{q}=500 \mathrm{~J}$
UPTU/UPSEE-2014
Thermodynamics
272730
Pressure-volume (PV) work done by an ideal gaseous system at constant volume is (where $E$ is internal energy of the system)
1 $-\Delta \mathrm{P} / \mathrm{P}$
2 zero
3 $-\mathrm{V} \Delta \mathrm{P}$
4 $-\Delta \mathrm{E}$
Explanation:
From the first law of thermodynamics $\Delta \mathrm{E}=\mathrm{q}+\mathrm{W}$ $\mathrm{W}=-\mathrm{P} \Delta \mathrm{V}$ $\therefore \Delta \mathrm{V}=0$ since the volume is constant $\mathrm{W}=0$
WB-JEE-2013
Thermodynamics
272660
Five moles of a gas are put through a series of changes as shown graphically in a cyclic process. The processes $A \rightarrow B, B \rightarrow C$ and $C \rightarrow$ A respectively are
1 isochoric, isobaric, isothermal
2 isobaric, isochoric, isothermal
3 isothermal, isobaric, isochoric
4 isochoric, isothermal, isobaric
Explanation:
Process $\mathrm{A} \rightarrow \mathrm{B}$ is isochoric, i.e., volume remains constant.
Process $\mathrm{B} \rightarrow \mathrm{C}$ is isobaric, i.e., pressure remains constant. ( $\because$ at constant $P, V \propto T$ )
Process $\mathrm{C} \rightarrow \mathrm{A}$ is isothermal, i.e., temperature remains constant.
