JEE Main 2016 (Online) 9th April Morning Slot | Gaseous State Question 51 | Chemistry

This chapter is currently out of syllabus

PREVIOUS NEXT

JEE Main 2016 (Online) 9th April Morning Slot

MCQ (Single Correct Answer)

-1

Out of Syllabus

At very high pressures, the compressibility factor of one mole of a gas is given by :

$${{pb} \over {RT}}$$

1 + $${{pb} \over {RT}}$$

1 $$-$$ $${{pb} \over {RT}}$$

1 $$-$$ $${b \over {\left( {VRT} \right)}}$$

JEE Main 2016 (Offline)

MCQ (Single Correct Answer)

-1

Out of Syllabus

Two closed bulbs of equal volume $$(V)$$ containing an ideal gas initially at pressure $${p_i}$$ and temperature $${T_1}$$ are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to $${T_2}.$$ The final pressure $${p_f}$$ is :

JEE Main 2016 (Offline) Chemistry - Gaseous State Question 52 English

JEE Main 2016 (Offline) Chemistry - Gaseous State Question 52 English

$$2{p_i}\left( {{{{T_2}} \over {{T_1} + {T_2}}}} \right)$$

$$2{p_i}\left( {{{{T_1}{T_2}} \over {{T_1} + {T_2}}}} \right)$$

$${p_i}\left( {{{{T_1}{T_2}} \over {{T_1} + {T_2}}}} \right)$$

$$2{p_i}\left( {{{{T_1}} \over {{T_1} + {T_2}}}} \right)$$

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)

-1

Out of Syllabus

If Z is a compressibility factor, van der Waals equation at low pressure can be written as:

Z = 1 + $$RT \over Pb$$

Z = 1 - $$a \over VRT$$

Z = 1 - $$Pb \over RT$$

Z = 1 + $$Pb \over RT$$

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)

-1

Out of Syllabus

For gaseous state, if most probable speed is denoted by C*, average speed by $$\mathop C\limits^{\_\_} $$ and mean square speed by C, then for a large number of molecules the ratios of these speeds are:

C*: $$\mathop C\limits^{\_\_} $$ : C = 1.128 : 1.225 : 1

C*: $$\mathop C\limits^{\_\_} $$ : C = 1.225 : 1.128 : 1

C*: $$\mathop C\limits^{\_\_} $$ : C = 1 : 1.225 : 1.128

C*: $$\mathop C\limits^{\_\_} $$ : C = 1 : 1.128 : 1.225

PREVIOUS NEXT

JEE Main Subjects