Advertisements
Advertisements
प्रश्न
Refer to figure. Let ∆U1 and ∆U2 be the changes in internal energy of the system in the process A and B. Then _____________ .

विकल्प
∆U1 > ∆U2
∆U1 = ∆U2
∆U1 < ∆U2
∆U1 ≠ ∆U2
Advertisements
उत्तर
∆U1 = ∆U2
The internal energy of the system is a state function, i.e. it only depends on the initial and final point of the process and doesn't depend on the path followed. Both processes A and B have common initial and final points. Therefore, change in internal energy in process A is equal to the change in internal energy in process B. Thus,
∆U1 = ∆U2 = 0
APPEARS IN
संबंधित प्रश्न
Explain why Two bodies at different temperatures T1 and T2, if brought in thermal contact, do not necessarily settle to the mean temperature (T1 + T2)/2.
Should the internal energy of a system necessarily increase if its temperature is increased?
When we rub our hands they become warm. Have we supplied heat to the hands?
The final volume of a system is equal to the initial volume in a certain process. Is the work done by the system necessarily zero? Is it necessarily nonzero?
Figure shows two processes A and B on a system. Let ∆Q1 and ∆Q2 be the heat given to the system in processes A and B respectively. Then ____________ .

An ideal gas goes from the state i to the state f as shown in figure. The work done by the gas during the process ______________ .

The pressure p and volume V of an ideal gas both increase in a process.
(a) Such a process is not possible.
(b) The work done by the system is positive.
(c) The temperature of the system must increase.
(d) Heat supplied to the gas is equal to the change in internal energy.
Figure shows a cylindrical tube of volume V with adiabatic walls containing an ideal gas. The internal energy of this ideal gas is given by 1.5 nRT. The tube is divided into two equal parts by a fixed diathermic wall. Initially, the pressure and the temperature are p1, T1 on the left and p2, T2 on the right. The system is left for sufficient time so that the temperature becomes equal on the two sides. (a) How much work has been done by the gas on the left part? (b) Find the final pressures on the two sides. (c) Find the final equilibrium temperature. (d) How much heat has flown from the gas on the right to the gas on the left?

A mixture of fuel and oxygen is burned in a constant-volume chamber surrounded by a water bath. It was noticed that the temperature of water is increased during the process. Treating the mixture of fuel and oxygen as the system,
- Has heat been transferred?
- Has work been done?
- What is the sign of ∆U?
A system releases 130 kJ of heat while 109 kJ of work is done on the system. Calculate the change in internal energy.
A system releases 100 kJ of heat while 80 kJ of work is done on the system. Calculate the change in internal energy.
Explain given cases related to energy transfer between the system and surrounding –
- energy transferred (Q) > 0
- energy transferred (Q) < 0
- energy transferred (Q) = 0
One gram of water (1 cm3) becomes 1671 cm3 of steam at a pressure of 1 atm. The latent heat of vaporization at this pressure is 2256 J/g. Calculate the external work and the increase in internal energy.
An ideal gas is compressed at a constant temperature. Its internal energy will ____________.
Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following:
What is the change in internal energy of the gas?
Figure shows the P-V diagram of an ideal gas undergoing a change of state from A to B. Four different parts I, II, III and IV as shown in the figure may lead to the same change of state.

- Change in internal energy is same in IV and III cases, but not in I and II.
- Change in internal energy is same in all the four cases.
- Work done is maximum in case I
- Work done is minimum in case II.
In thermodynamics, heat and work are ______.
If a gas is compressed adiabatically:
The molar specific heat of He at constant volume is 12.47 J/mol.K. Two moles of He are heated at constant pressure. So the rise in temperature is 10 K. Find the increase in internal energy of the gas.
A system releases 125 kJ of heat while 104 kJ work is done on the system. Calculate the change in internal energy.
