NEET Practice Questions (MCQs) with Answers & Solutions

For this problem, we need to determine the values of $q$, $w$, and $ riangle E$.

Given data:

• $q = 250$ J (heat absorbed by the gas)
• Initial volume $V_i = 1.0$ L
• Final volume $V_f = 10.0$ L
• External pressure $P = 0.50$ atm
• 1 L atm = 101 J

First, calculate the work done by the gas during expansion:

$w = -P riangle V = -P (V_f - V_i) = -0.50$ atm $(10.0 - 1.0)$ L

$w = -0.50 imes 9.0$ L atm = $-4.5$ L atm

Convert L atm to Joules:

$w = -4.5 imes 101$ J = $-454.5$ J

Rounding, $w ightarrow -455$ J

Next, calculate the change in internal energy, $ riangle E$:

$ riangle E = q + w = 250$ J $+ (-455)$ J = $-205$ J

Thus, the values are: $q = 250$ J $w = -455$ J $ riangle E = -205$ J

To find the enthalpy of formation ($ riangle H_f^ ext{°}$) of $H_2O_2(l)$, we can use the given reaction and enthalpy values. The given data is:

$ riangle H^ ext{°}$ for the reaction $H_2O_2(l) ightarrow H_2O(l) + rac{1}{2}O_2(g) = -23.5$ kcal/mol

$ riangle H_f^ ext{°}(H_2O(l)) = -68.3$ kcal/mol

We need to find $ riangle H_f^ ext{°}(H_2O_2(l))$.

From the reaction, we can write:

$ riangle H^ ext{°} = riangle H_f^ ext{°}(H_2O(l)) + rac{1}{2} riangle H_f^ ext{°}(O_2(g)) - riangle H_f^ ext{°}(H_2O_2(l))$

Since $ riangle H_f^ ext{°}(O_2(g)) = 0$ (standard state), we have:

$-23.5 = -68.3 - riangle H_f^ ext{°}(H_2O_2(l))$

Rearranging to solve for $ riangle H_f^ ext{°}(H_2O_2(l))$:

$ riangle H_f^ ext{°}(H_2O_2(l)) = -68.3 + 23.5 = -44.8$ kcal/mol

Thus, the enthalpy of formation of $H_2O_2(l)$ is $-44.8$ kcal/mol.

In a cyclic process, the change in internal energy ( ext{ΔU}) of the system is zero because the system returns to its initial state. According to the first law of thermodynamics, ext{ΔU = Q - W}, where Q is the heat absorbed by the system and W is the work done by the system. Given that ext{ΔU} is zero for a cyclic process, the equation simplifies to ext{Q = W}. Since the work done by the system is -50 kJ, the heat absorbed by the system must also be -50 kJ. However, since the question asks for the heat absorbed, we consider the positive value of the work done by the system (since it is energy taken out of the system), which is 50 kJ.

$ C_v = {3 \over 2} RT ; C_p = { 5 \over 2} RT $ for monoatomic gas $ C_v = { 5 \over 2} RT ; C_p = { 7 \over 2} RT $ for diatomic gas Thus for mixture of 1 mole each, $ C_v = { { 3\over 2 } RT + { 5 \over 2} RT \over 2 } and C_p= { { 5 \over 2 } RT + {7 \over 2} RT \over 2 } $ Therefore $ C_p /C_v = { 3RT \over 2 RT } = 1.5 $

78g of benzene on combustion produces heat = – 3264.6 kJ $ \therefore 39g will produce = { - 3264.6 \over 2} = - 1632.3 kJ.$

First, calculate the total energy that can be provided by the 11.2 kg of butane. The molar mass of butane ( ext{C_4H_{10}}) is ext{58 g/mol}, so the number of moles in 11.2 kg is ext{11200 g / 58 g/mol = 193.1 mol}. The energy released per mole of butane is -2658 kJ, so the total energy available is ext{193.1 mol * 2658 kJ/mol = 513,679.8 kJ}. Given that a normal family needs 20000 kJ of energy per day, the number of days the cylinder will last is ext{513,679.8 kJ / 20000 kJ/day ≈ 25.68 days}. Since we are looking for an integer value, the closest option is 26 days.

The reaction given is the transition of carbon from diamond to graphite with a negative enthalpy change (ΔH = -453.5 cal). A negative ΔH indicates that the process releases heat, meaning the product (graphite) is more stable than the reactant (diamond). Hence, graphite is more stable than diamond.

In terms of energy units, the correct order is: $1 ext{ cal} > 1 ext{ joule} > 1 ext{ erg}$. Specifically, $1 ext{ cal} = 4.184 ext{ joules}$ and $1 ext{ joule} = 10^7 ext{ ergs}$. Therefore, $1 ext{ cal} > 1 ext{ joule} > 1 ext{ erg}$.

For 2 moles of water vapour, Absorbed energy by system is $ \triangle H_{vap} = 2 \times 9720 = 19440 cal $ $ \triangle S _ {vap} = { \triangle H_{vap} \over T_b } $ $ = { 19440 \over (100 + 273) } $ $ = 52.12 Cal. K^ {-1} mole ^ {-1} $ $ = 52.12 4.184 $ $ = 217.6 joule K ^ {-1} . mole ^ {-1} $