The one subject in NEET which is candidates who can easily attain good marks is Chemistry. That's the reason, often, one doesn’t pay notice and choose to compromise it. But if one wants to rank above others, the tip is to be thorough with NEET chemistry concepts. The understanding of reactions and definite basic understanding is what requires major attention in Chemistry but once done it only gets simpler from there. The main focus on the to-do list should be on getting a hang of the NCERT syllabus of NEET chemistry..
Q1. Given,
Change in entropy is negative for:
Solution
(d) The gaseous phase have more entropy and thus, ∆S is +ve in (a) and (b). Also decrease in pressure increases disorder and thus, ∆S is +ve in (c). In (d) the disorder decreases in liquid state due to decrease in temperature. Thus, ∆S=-ve.
(d) The gaseous phase have more entropy and thus, ∆S is +ve in (a) and (b). Also decrease in pressure increases disorder and thus, ∆S is +ve in (c). In (d) the disorder decreases in liquid state due to decrease in temperature. Thus, ∆S=-ve.
Q2.When 1 mole of gas is heated at constant volume. Temperature is raised from 298 to 308 K. Heat supplied to the gas is
500 J. Then which statement is correct?
500 J. Then which statement is correct?
Solution
(b) At constant volume P∆V=0,∴q=∆U.
(b) At constant volume P∆V=0,∴q=∆U.
Q3. The standard heat of formation of sodium ions in aqueous solution from the following data will be:
Heat of solution NaOH(aq.) from NaOH(s)= -470.7 kJ Heat of formation of OH- (aq.) From OH- (s)= -228.8 kJ
Heat of solution NaOH(aq.) from NaOH(s)= -470.7 kJ Heat of formation of OH- (aq.) From OH- (s)= -228.8 kJ
Solution
(c) Na+(s)+OH-(s)+aq.→Na+(aq.)+OH-(aq.); ∆H= -470.7 kJ OH-(s)+aq.→OH-(aq.); ∆H= -228.8 kJ ∴ Na+(s)+aq.→Na+(aq.); ∆H= -241.9 kJ
(c) Na+(s)+OH-(s)+aq.→Na+(aq.)+OH-(aq.); ∆H= -470.7 kJ OH-(s)+aq.→OH-(aq.); ∆H= -228.8 kJ ∴ Na+(s)+aq.→Na+(aq.); ∆H= -241.9 kJ
Q4. Internal energy is
Solution
(a) The exact value of internal energy is not known as it includes all type of energies of molecules constituting the given mass of matter such as translational, vibrational and rotational. The kinetic and potential energy of the nuclei and electron with in the individual molecules and the manner in which the molecule are linked together, are E=E_translational+E_rotational+E_vibrational Thus, we can say that internal energy is partly potential ad partly kinetic
(a) The exact value of internal energy is not known as it includes all type of energies of molecules constituting the given mass of matter such as translational, vibrational and rotational. The kinetic and potential energy of the nuclei and electron with in the individual molecules and the manner in which the molecule are linked together, are E=E_translational+E_rotational+E_vibrational Thus, we can say that internal energy is partly potential ad partly kinetic
Q5.S+3/2 O2 → SO3 + 2x kcal;
SO2 + 1/2 O2 → SO3 + y kcal; The heat of formation of SO2 is :
SO2 + 1/2 O2 → SO3 + y kcal; The heat of formation of SO2 is :
Solution
a) S+3/2 O2→SO3; ∆H= -2x SO2+1/2 O2→CO2+SO3; ∆H= -y S+O2→S2; ∆H=(y-2x)
a) S+3/2 O2→SO3; ∆H= -2x SO2+1/2 O2→CO2+SO3; ∆H= -y S+O2→S2; ∆H=(y-2x)
Q6. Any series of operation so carried out that at the end, the system is back to its state is called
Solution
(D)
(D)
Q7.An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If Ti is the initial temperature
and Tf is the final temperature, which of the following statements is correct?
and Tf is the final temperature, which of the following statements is correct?
Solution
(a) (Tf)irreversible>(Tf)reversible it is an adiabatic expansion and W(rev) is maximum.
(a) (Tf)irreversible>(Tf)reversible it is an adiabatic expansion and W(rev) is maximum.
Q8.The work done by a weightless piston in causing an expansion ∆V (at constant temperature), when the opposing
pressure, P is variable, is given by:
pressure, P is variable, is given by:
Solution
a) Wrev=-∫PdV or-∫P∆V; note that opposing pressure is not constant throughout
a) Wrev=-∫PdV or-∫P∆V; note that opposing pressure is not constant throughout
Q9. In an irreversible process, the value of ∆Ssystem + ∆Ssurr is:
Solution
a) ∆Stotal=∆S(sys.)+∆S(surr.)=+ve.
a) ∆Stotal=∆S(sys.)+∆S(surr.)=+ve.
Q10. Three moles of an ideal gas expanded spontaneously into vacuum. The work done will be:
Solution
(d) W=P×∆V =0×∆V=0
(d) W=P×∆V =0×∆V=0