A. Deviate from ideal gas behaviour to the same degree
B. Have nearly the same compressibility factor
C. Both A. and B.
D. Neither A. nor B.
Advertisement
Related Mcqs:
- Under conditions of equal reduced pressure and equal reduced temperature, substances are said to be in the ‘corresponding states’. At equal reduced conditions i.e., at the corresponding state, the ____________ of different gases are nearly the same?
- A. Compressibility B. Molecular weight C. Humidity D. None of these...
- If two gases have same reduced temperature and reduced pressure, then they will have the same____________________?
- A. Volume B. Mass C. Critical temperature D. None of these...
- At higher temperatures, molal heat capacities for most of the gases (at constant pressure) _______________ with increase in temperature?
- A. Varies linearly B. Increases C. Decreases D. Does not vary...
- Near their critical temperatures, all gases occupy volumes _____________ that of the ideal gas?
- A. Less than B. Same as C. More than D. Half...
- With increase in reduced temperature, the fugacity co-efficient of a gas at constant reduced pressure___________________?
- A. Increases B. Decreases C. Remain same D. Decreases linearly...
- Compressibility factor-reduced pressure plot on reduced co-ordinates facilitates__________________?
- A. Use of only one graph for all gases B. Covering of wide range C. Easier plotting D. More accurate plotting...
- Except for monatomic gases, the molal heat capacity at constant volume for all gases is _____________ Kcal/Kg mole.° K?
- A. 3 B. > 3 C. < 3 D. < 1...
- Which of the following gases is having the widest explosion limit (about 2 to 81% gas in gasair mixture), rendering it the property of the most explosive gas ?
- A. Hydrogen B. Acetylene C. Carbon monoxide D. Ammonia...
- The equilibrium constant for a chemical reaction at two different temperatures is given by__________________?
- A. Kp2/Kp1 = – (ΔH/R) (1/T2 – 1/T1) B. Kp2/Kp1 = (ΔH/R) (1/T2 – 1/T1) C. Kp2/Kp1 = ΔH (1/T2 – 1/T1) D. Kp2/Kp1 = – (1/R) (1/T2 – 1/T1)...
- What is the value of maximum COP in case of absorption refrigeration, if refrigeration provided is at temperature, TR (where, T1 and T2 are source & surrounding temperatures respectively.) ?
- A. TR/(T2 – TR) × (T1 – T2)/T1 B. TR/(T2 – TR) × T1/(T1 – T2) C. TR/(T1 – TR) × (T1 – T2)/T1 D. None of these...
Advertisement