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

Two moles of an ideal gas are expanded isothermally from $15 {dm}^{3}$ to $20 {dm}^{3}$ . If the amount of work done is $- 6 {dm}^{- 3}$ bar, find external pressure needed to obtain this work.

A. $1.2 \times 10^{5} Pa$

B. $3.2 Pa$

C. $8.1 \times 10^{4} Pa$

D. $2.4 Pa$

Correct Option is (A)

$\begin{aligned} V_{1} = 15 {dm}^{3}, V_{2} = 20 {dm}^{3}, W = - 6 {dm}^{3} bar \\ W = - P_{ext} Δ V \\ ∴ P_{ext} = - \frac{W}{Δ V} = - \frac{W}{(V_{2} - V_{1})} \\ = \frac{- (- 6 {dm}^{3} bar)}{(20 - 15) {dm}^{3}} = \frac{6}{5} bar \\ = 1.2 bar = 1.2 \times 10^{5} Pa \end{aligned}$

7. ⇒ (MHT CET 2021 21th September Morning Shift )

A gas is allowed to expand in an insulated container against a constant external pressure of 2.5 atm from $2.5 L$ to $4.5 L$ , the change in internal energy of the gas in joules is

A. $- 836.3 J$

B. $- 1136.2 J$

C. $- 450 J$

D. $- 506.5 J$

Correct Option is (D)

The change in internal energy is given by, $Δ U = Q + W$

As container is insulated, $Q = 0$

$\begin{aligned} ∴ Δ U & = W = - P_{ext} Δ V = - P_{ex} (V_{2} - V_{1}) \\ ∴ Δ U & = - 2.5 atm (4.5 L - 2.5 L) \\ = - 5.0 L . atm \\ = - 5.0 \times 101.325 J \\ ∴ Δ U & = - 506.625 J \end{aligned}$

8. ⇒ (MHT CET 2021 21th September Morning Shift )

What is the work done when a gas is compressed from 2.5 $\times$ 10 $^{- 2}$ m $^{3}$ to 1.3 $\times$ 10 $^{- 2}$ m $^{3}$ at constant external pressure of 4.05 bar?

A. 4050 J

B. 4400 J

C. 4200 J

D. 4860 J

Correct Option is (D)

$\begin{aligned} V_{1} = 2.5 \times 10^{- 2} m^{3} = 25 {dm}^{3}, V_{2} = 1.3 \times 10^{- 2} m^{3} = 13 {dm}^{3} \\ \begin{aligned} P_{ext} = & 4.05 bar, W = ? \\ W & = - P_{ext} Δ V = - P_{ext} (V_{2} - V_{1}) \\ = - 4.05 bar \times (13 - 25) {dm}^{3} \\ = 48.6 {dm}^{3} bar \\ = 48.6 \times 100 J \\ ∴ W & = 4860 J \end{aligned} \end{aligned}$

9. ⇒ (MHT CET 2021 20th September Morning Shift )

An ideal gas on isothermal reversible compression from 10L to 5L performs 1730J of work at 300 K. Calculate number of moles of gas involved in compression? (R = 8.314 J K $^{- 1}$ mol $^{- 1}$ )

A. 2.5

B. 2

C. 1

D. 3

Correct Option is (C)

$V_{1} = 10 L, V_{2} = 5 L, W = 1730 J, T = 300 K, R = 8.314 J K^{- 1} {mol}^{- 1}, n = ?$

$W_{max} = - 2.303 nRT \log_{10} \cdot \frac{V_{2}}{V_{1}}$

$∴ 1730 J = - 2.303 n \times 8.314 {JK}^{- 1} {mol}^{- 1} \times 300 K \times \log_{10} (\frac{5}{10})$

$\begin{aligned} ∴ n = \frac{1730 J}{- 2.303 \times 8.314 J K^{- 1} {mol}^{- 1} \times 300 K \times \log_{10} (0.5)} \\ ∴ n = \frac{1730}{- 5744.14 \times (- 0.3010)} mol \\ ∴ n = \frac{1730}{1729} \approx 1 mol \end{aligned}$

⇒Thermodynamics