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n moles of an ideal monoatomic gas is kept in a closed vessel of volume 0.0083 \mathrm{~m}^3at a temperature of 300K and a pressure of 1.6 \times 10^6 \mathrm{~Pa}. Heat energy of 2.49 \times 10^4 \mathrm{~J} is supplied to the gas. Given R=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}

Option: 1

The value of n = 5


Option: 2

\mathrm{\text { For the gas } C_p=20.75 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}}


Option: 3

\mathrm{\text { The final temperature of the gas is } 302^{\circ} \mathrm{C}}


Option: 4

\mathrm{\text { The Final pressure of the gas is } 3.6 \times 10^5 \mathrm{~Pa}}


Answers (1)

best_answer

\mathrm{\begin{aligned} & \text { (a) } P V=n R T \Rightarrow n=\frac{P V}{R T} \\ & =\frac{\left(1.6 \times 10^6\right) \times 0.0083}{8.3 \times 300}=\frac{16}{3} \end{aligned}}

\mathrm{\text { (b) For a monoatomic gas } C_p=\frac{5 R}{2}=\frac{5}{2} \times 8.3}

\mathrm{=20.75 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}}

\mathrm{\begin{aligned} & \text { (c) } \Delta Q=n C_v \Delta T ; C_v=\frac{3 R}{2} \\ & \therefore \Delta T=\frac{\Delta Q}{n C_v}=\frac{2 \Delta Q}{3 n R}=\frac{2 \times 2.49 \times 10^4}{3 \times \frac{16}{3} \times 8.3} \\ & =375 \mathrm{~K} \end{aligned}}

∴ Final temperature = 300 + 375 = 675 K

\mathrm{\begin{aligned} & \text { (d) } \frac{P_2}{T_2}=\frac{P_1}{T_1} \Rightarrow P_2 \frac{P_1 \times T_2}{T_1}=\frac{1.6 \times 10^6 \times 675}{300} \\ & =3.6 \times 10^6 \mathrm{~Pa} \end{aligned}}

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seema garhwal

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