These are Assertion-Reason type questions.Question contains two statements:Statement-1 (Assertion) and Statement-2 (Reason). Answer these question from the following four options.

Statement 1 :The mass and volume of a substance are the extensive properties and are proportional to each other.
Statement 2 :The ratio of mass of a sample to its volume is an intensive property

Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1
Statement-1 is True, Statement-2 is True Statement-2 is NOT a correct explanation for Statement-1
Statement - 1 is True, Statement-2 is False
Statement -1 is False, Statement-2 is True

Solution

The mass and volume depend upon the quantity of matter so these are extensive properties while ratio of mass to its volume does not depend upon the quantity of matter so this ratio is an extensive property.

For vaporization of water at 1 atmospheric pressure, the values of ?H and ?S are 40.63 kJ mol^-1 and 108.8 JK^-1 mol^-1,respectively. The temperature when Gibbs energy change (?G) for this transformation will be zero, is:

293.4K
273.4 K
393.4K
373.4 K

Solution

Which of the following are not state functions ?
(I)q + w
(II)q
(III)w
(IV)H – TS

(I) and (IV)
(II), (III) and (IV)
(I), (II) and (III)
(II) and (III)

Solution

We know that heat (q) and work (w) are not state functions but (q + w) is a state function. H – TS (i.e. G) is also a state function. Thus II and III are not state functions so the correct answer is option (d).

Given that bond energies of H – H and Cl –Cl are 430 kJ mol^-1and 240 kJ mol^-1 respectively and ?_fH for HCl is – 90 kJ mol^-1, bond enthalpy of HCl is

380 kJ mol^-1
425 kJ mol^-1
245 kJ mol^-1
290 kJ mol^-1

Solution

Solution

The heat of atomization of PH₃(g) is 228 kcal mol^-1 and that of P₂H₄(g) is 335 kcal mol^-1. The energy of the P–P bond is

102 kcal mol^-1
51 kcal mol^-1
26 kcal mol^-1
204 kcal mol^-1

Solution

One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27°C. If the work done during the process is 3 kJ, then final temperature of the gas is(C_v= 20 J/K):

100 K
150 K
195 K
255 K

Solution

Since the gas expands adiabatically (i.e., no change in enthalpy) so the heat is totally converted into work.For the gas CV= 20 J/K. Thus, 20 J of heat is required for 1° change in temperature of the gas.Heat change involved during the process (i.e., work done)= 3 kJ = 3000 J

Change in temperature \(\frac{300}{20}\)k = 150 K
Initial temperature = 300 K
Since the gas expands so the temperature decreases and thus, final temperature is 150 K (300 – 150 = 150).