6. ⇒ (MHT CET 2023 10th May Morning Shift )

What is change in internal energy when system releases $8 kJ$ of heat and performs $660 J$ of work on the surrounding?

A. $- 7340 J$

B. $- 5400 J$

C. $- 8660 J$

D. $- 1212 J$

Correct Option is (C)

$Δ U = Q + W = - 8000 J - 660 J = - 8660 J$

When heat is released by the system to the surroundings, Q is negative. When work is done by the system on the surroundings, W is negative.

7. ⇒ (MHT CET 2023 9th May Evening Shift )

What is the value of $Δ H - Δ U$ for the following reaction?

$2 C_{(s)} + 3 H_{2 (g)} \to C_{2} H_{6 (g)}$

A. $4 RT$

B. $- 5 RT$

C. $RT$

D. $- 2 RT$

Correct Option is (D)

$\begin{aligned} Δ n_{g} = (moles of product gases) - (moles of reactant gases) \\ Δ n_{g} = 1 - 3 = - 2 mol \\ Now, using formula, \\ Δ H = Δ U + Δ n_{g} RT \\ ∴ & Δ H = Δ U - 2 RT \\ ∴ & Δ H - Δ U = - 2 RT \end{aligned}$

8. ⇒ (MHT CET 2023 9th May Morning Shift )

An ideal gas expands by performing $200 J$ of work, during this internal energy increases by $432 J$ . What is enthalpy change?

A. $200 J$

B. $232 J$

C. $432 J$

D. $632 J$

Correct Option is (D)

Assuming constant pressure,

$Δ H = Δ U + P_{ext} Δ V = Δ U - W$

Work done $(W) = - 200 J$ (for expansion of a gas)

$\begin{aligned} Δ U = + 432 J \\ Δ H = Δ U - W = + 432 - (- 200) = + 632 J \end{aligned}$

9. ⇒ (MHT CET 2021 21th September Evening Shift )

When 1 mole of gas is heated at constant volume, the temperature rises form $273 K$ to $546 K$ . If heat supplied to the gas is $x J$ , then find the correct statement from following.

A. $Q = Δ U = x J, W = 0$

B. $Q = W = x J, Δ V = 0$

C. $Δ V = 0, Q = W = - x J$

D. $Q = - W = x J, Δ V = 0$

Correct Option is (A)

At constant volume, $Δ V = 0$

$∴ W = - P Δ V = 0$

According to first law of thermodynamics,

$\begin{aligned} Δ U & = Q + W \\ ∴ Δ U & = Q = x J \end{aligned}$

10. ⇒ (MHT CET 2021 21th September Evening Shift )

What is the value of $Δ H - Δ U$ for the formation of 2 moles of ammonia from $H_{2 (g)}$ and $N_{2 (g)}$ ?

A. $- \frac{RT}{2}$

B. $\frac{RT}{2}$

C. $-$ 2RT

D. 2RT

Correct Option is (C)

$\begin{matrix} N_{2 (g)} + 3 H_{2 (g)} ⟶ 2 {NH}_{3 (g)} \\ Δ n_{(g)} = n_{2} - n_{1} = 2 - 4 = - 2 \\ Δ H - Δ U = Δ n_{(g)} RT = - 2 RT \end{matrix}$

⇒Thermodynamics