
A gas absorbs 250J of heat and expands from 1litre to 10litre at constant temperature against external pressure of 0.5atm. The values of q, w and $\Delta U$ will be respectively:
(A) -250J, 455J, 710J
(B) 250J, -455J, -205J
(C) -250J, -455J, -205J
(D) -250J, 455J, 205J
Answer
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Hint: Recollect the basic concepts of thermodynamics like work done and sign conventions when work is done on the system and when work is done by the system as well as heat is absorbed by the system or when heat is released by the system. Think about the mathematical expression of thermodynamics and solve it to get the correct answer.
Complete step by step answer:
- In thermodynamics, there are sign conventions to indicate the energy that is taken into the system and that is given out by the system. By tradition there are four sign conventions:
+q: Heat is absorbed by the system from the surroundings.
-q: Heat is released by the system into the surroundings.
+w: Work is done on the system by the surroundings.
-w: Work is done by the system on the surroundings.
- In this problem, a gas absorbs 250J of heat and therefore, q = +250J
- Work done by the system is given as, $w=-{{p}_{ex}}.\Delta V$ where w is the work done, ${{p}_{ex}}$ is the external pressure and $\Delta V$ is the change in volume.
- Therefore, $w=-0.5\times (10-1)=-0.5\times 9=-4.55L.atm$. We need to find work done in joules and 1L.atm=101.3J. Therefore we get, $w=-4.55\times 101.3=-455J$
- Now, we know the mathematical expression for the first law of thermodynamics that is, $\Delta U=q+w$ where $\Delta U$ is the change in internal energy, q is the heat and w is the work done.
- Therefore, change in internal energy, $\Delta U=+250J-455J=-205J$
- Therefore, the values of q, w and $\Delta U$ will be 250J, -455J and -205J respectively.
So, the correct answer is “Option B”.
Note: Remember the sign conventions in thermodynamics. First law of thermodynamics tells us that the change in internal energy of the system is the sum of heat energy supplied to the system and work done on the system by the surroundings.
Complete step by step answer:
- In thermodynamics, there are sign conventions to indicate the energy that is taken into the system and that is given out by the system. By tradition there are four sign conventions:
+q: Heat is absorbed by the system from the surroundings.
-q: Heat is released by the system into the surroundings.
+w: Work is done on the system by the surroundings.
-w: Work is done by the system on the surroundings.
- In this problem, a gas absorbs 250J of heat and therefore, q = +250J
- Work done by the system is given as, $w=-{{p}_{ex}}.\Delta V$ where w is the work done, ${{p}_{ex}}$ is the external pressure and $\Delta V$ is the change in volume.
- Therefore, $w=-0.5\times (10-1)=-0.5\times 9=-4.55L.atm$. We need to find work done in joules and 1L.atm=101.3J. Therefore we get, $w=-4.55\times 101.3=-455J$
- Now, we know the mathematical expression for the first law of thermodynamics that is, $\Delta U=q+w$ where $\Delta U$ is the change in internal energy, q is the heat and w is the work done.
- Therefore, change in internal energy, $\Delta U=+250J-455J=-205J$
- Therefore, the values of q, w and $\Delta U$ will be 250J, -455J and -205J respectively.
So, the correct answer is “Option B”.
Note: Remember the sign conventions in thermodynamics. First law of thermodynamics tells us that the change in internal energy of the system is the sum of heat energy supplied to the system and work done on the system by the surroundings.
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