Thermodynamic Processes
Trending Questions
Q.
At 27∘ C, the ratio of rms velocities of ozone to oxygen is
√35
√43
√23
0.25
Q. What is the work which could be obtained from an isothermal reversible expansion of 1 mol of Cl2 from 1 dm3 to 50 dm3 at 273 K using ideal gas behaviour.
- −8.88 kJ mol−1
- 8.88 kJ mol−1
- −88.8 kJ mol−1
- 88.8 kJ mol−1
Q. In which of the following processes does the entropy decrease?
- Conversion of oxygen to ozone
- Conversion of dry ice to CO2(g)
- Thermal decomposition of sodium bicarbonate
- Desorption of H2(g) from nickel surface
Q.
Match the thermodynamic processes given below:
Column IColumn IIA. Freezing of water at 273K and 1 atmp. q=0B. Expansion of 1 mole of an ideal gas into a vacuum under isolated conditions q. W=0C. Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated containerr. ΔSsys<0D. Reversible heating of H2(g) at 1 atm from 300 K to 600K, followed by reversible cooling to 300K at 1 atms. ΔU=0t. ΔG=0
(A) →r, t; (B) →q, s, t; (C) →p, q, s; (D) →p, q, s
(A) →r, t; (B) →p, q, s; (C) →p, q, s; (D) →q, s, t
(A); →q, s, t(B) →p, q, s; (C) →p, q, s; (D) →r, t
(A) →r, t; (B) →p, q, s; (C) →p, q, s; (D) →q, s, t
Q. In an adiabatic process
- q=0
- ΔT=0
- ΔP=0
- ΔV=0
Q. A fixed mass 'm' of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the figure.
The succeeding operations that enable this transformation of states are:
The succeeding operations that enable this transformation of states are:
- heating, cooling, heating, cooling
- cooling, heating, cooling, heating
- heating, cooling, cooling, heating
- cooling, heating, heating, cooling
Q. 160 g of oxygen gas expand at STP to occupy double of its original volume. The work during the process is:
(1 L.atm=101.3J)
(1 L.atm=101.3J)
- −4.7 kcal
- −2.7 kcal
- −6.7 kcal
- −8.7 kcal
Q. The value of PV for 5.6 litres of an ideal gas at NTP is
- 0.35 RT
- 0.25 RT
- 0.30 RT
- 0.20 RT
Q. A piston filled with 0.04 mol of an ideal gas expands reversibly form 50.0 mL to 375 mL at a constant temperature of 37.0oC. As it does so, it absorbs 208 J of heat. The values of q and w for the process will be:
(R = 8.314 J/mol K) (ln 7.5 = 2.01)
(R = 8.314 J/mol K) (ln 7.5 = 2.01)
- q = -208 J, w = -208 J
- q = -208 J, w = +208 J
- q = +208 J, w = +208 J
- q = +208 J, w = -208 J
Q. One mole of an ideal monoatomic gas at temperature T and volume 3 L expands to 6 L against a constant external pressure of 2 atm under adiabatic conditions. Then the final temperature of the gas will be:
- T+302R
- T−123R
- T+203R
- T253+1
Q. 160 g of oxygen gas expand at STP to occupy double of its original volume. The work during the process is:
(1 L.atm=101.3J)
(1 L.atm=101.3J)
- −4.7 kcal
- −2.7 kcal
- −6.7 kcal
- −8.7 kcal