First Law of Thermodynamics
Trending Questions
Q.
State and explain the first law of thermodynamics?
Q.
What is Boyles temperature?
Q.
In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
Q. A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be
- – 500J
- – 505J
- + 505J
- 1136.25J
Q. The temperature of the system decreases in an:
- Adiabatic compression
- Isothermal compression
- Isothermal expansion
- Adiabatic expansion
Q. A piston filled with 0.04 mole of an ideal gas expand reversibly from 50.0mL to 375 mL at a consumed temperature of 37.0∘C. As it does so, it absorbs 208∘C heat. The values of q and W for the process will be
(R = 8.314J/mol K, In 7.5 = 2.01)
(R = 8.314J/mol K, In 7.5 = 2.01)
- q = + 208J, W =-208
- q = -208J, W = -208
- q = +208J, W =+208J
- q = -208, W = +208
Q. One mol of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27oC. If the work done during the process is 3 kJ, then the final temperature of gas is:
Given: Cv for the gas = 20 J/K
Given: Cv for the gas = 20 J/K
- 100 K
- 150 K
- 195 K
- 255 K
Q. An ideal gas expands against a constant external pressure of 2.0 atmosphere from 20 litre to 40 litre and absorbs 10 kJ of heat from surrounding. What is the change in internal energy of the system? (Given : 1 atm litre =101.3 J)
- 4052 J
- 9940 J
- 5948 J
- 14052 J
Q. Heat given to a system is 35 J and work done by the system is 15 J. The change in the internal energy of the system will be:
- 50 J
- 20 J
- 30 J
- 40 J
Q.
The volume of a given mass of monatomic gas changes with temperature according to the relation . The work done when temperature changes by will be . The value of is ____. [universal gas constant]
Q. Which of the following is the correct equation
- ΔU=Q−W
- W=ΔU+Q
- ΔU=W+Q
- None of these
Q.
How does the law of the conservation of energy relate to the First Law of Thermodynamics.
Q. PCl5 dissociates as
PCl5(g)⇌PCl3(g)+Cl2(g)
5 moles of PCl5 are placed in a 200 litre vessel which contains 2 moles of N2 and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant Kp for the dissociation of PCl5 is _____ ×10−3. (nearest integer)
(Given: R=0.082 L atm K−1mol−1; Assume ideal gas behaviour)
PCl5(g)⇌PCl3(g)+Cl2(g)
5 moles of PCl5 are placed in a 200 litre vessel which contains 2 moles of N2 and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant Kp for the dissociation of PCl5 is _____ ×10−3. (nearest integer)
(Given: R=0.082 L atm K−1mol−1; Assume ideal gas behaviour)
Q. SO2 and CH4 are introduced in a vessel in the molar ratio 1:2 . The ratio of molecules of two gases present in the container when their rate of effusion becomes equal is :
- 1:2
- 2:1
- 4:1
- 1:4
Q. One mole of a monoatomic ideal gas expands adiabatically at initial temperature T against a constant external pressure of 1atm from 1 litre to 3 litres. The final temperature of the gas is
- T
- T−2(1.5×0.082)
- T3(5R/2)
- T+2(1.5×0.082)
Q.
A system is provided with 200cal of heat and the work done by the system on the surrounding is 40J. the internal energy
- Increases by 600J
- Decreases by 800J
- Increase by 800J
- Increases by 160J
Q. ΔU is equal to
- Isochoric work
- Isobaric work
- Adiabatic work
- Isothermal work
Q. A man takes a diet equivalent to 10000 kJ per day and does work, in expending his energy in all forms equivalent to 12500 KJ per day. If the energy lost was stored as sucrose (1632 kJ per 100 g), how many days should it takes to lose 2 kg of his weight? (Ignore water loss).
- 13
- 52
- 26
- 39
Q. A system does 200 J of work and at the same time absorbs 150 J of heat. The magnitude of the change in internal energy inJ is (Nearest integer)
Q. For water ΔvapH=41kJ mol−1 at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas the that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ___kJ mol−1.
[Use : R=8.3 J mol−1K−1]
[Use : R=8.3 J mol−1K−1]
Q. Energy equivalent to one erg, one joule and one calorie are in order:
- 1 erg > 1 J > 1 cal
- 1 erg > 1 cal > 1 J
- 1 cal > 1 J > 1 erg
- 1 J > 1 cal > 1 erg
Q. Find the work done in J when one mole of an ideal gas in 10 litre container at 1 atm is allowed to enter into an evacuated bulb of capacity 100 litre
Q. In an isochoric process the increase in internal energy is
- Equal to the heat absorbed
- Equal to the heat evolved
- Equal to the work done
Equal to the sum of the heat absorbed and work done
Q. A gas absorbs 120 J of heat and expands against the external pressure of 1.10 atm from a volume of 0.5 L to 2.0 L. What is the change in internal energy ? (1 atm = 101.3 J)
- -47.1 J
- -25.2 J
- -35.5 J
- 40.2 J
Q. In an isochoric process the increase in internal energy is
- Equal to the heat absorbed
- Equal to the heat evolved
- Equal to the work done
Equal to the sum of the heat absorbed and work done
Q. The thermodynamic process in which the temperature remains constant is:
- Adiabatic process
- Isobaric process
- Isothermal process
- Isochoric process
Q. The internal energy when a system goes from state A to B is 40 kJ. If the system goes from A to B by a reversible path and returns to state A by an irreversible path. What would be the net change in internal energy?
- > 40 kJ
- < 40 kJ
- 40 kJ
- zero
Q. A system absorbs 20 kJ heat and also does 10 kJ of work. The net internal energy of the system:
- Increases by 10 kJ
- Increases by 30 kJ
- Decreases by 10 kJ
- Decreases by 30 kJ
Q. In a cyclic process, heat transfers are +14.7 kJ , −25.2 kJ, −3.56 kJ and +31.5 kJ. What is the net work done in the net cyclic process?
- +10.56 kJ
- −10.56 kJ
- −17.44 kJ
- +17.44 kJ
Q. The work done in an open vessel at 300 K, when 112 g iron reacts with dil HCl is :
- 1200 cal
- 600 cal
- 200 cal
- 300 cal