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  • An ideal gas (γ = 1.4) expands from <img alt="5 \times 10^{-3} \mathrm{~m}^3 \text { to } 25 \times 10^{-3} \mathrm{~m}^3" src="https://entrancecorner.oncodecogs.com/gif.latex?5%20%5Ctimes%20

An ideal gas (γ = 1.4) expands from 5 \times 10^{-3} \mathrm{~m}^3 \text { to } 25 \times 10^{-3} \mathrm{~m}^3 at a constant pressure of 1 \times 10^5 \mathrm{~Pa}. The heat energy supplied to the gas in this process is

Option: 1

7 J


Option: 2

70 J


Option: 3

700 J


Option: 4

7000 J


Answers (1)

best_answer

Work done on the gas is

\mathrm{\begin{aligned} & \Delta W=P \Delta V=P\left(V_f-V_i\right) \\ & =1 \times 10^5 \times(25-5) \times 10^{-3} \\ & =2000 \mathrm{~J} \end{aligned}}

The internal energy is given by \mathrm{U=\frac{P V}{(\gamma-1)}}

\mathrm{\therefore U_i=\frac{P V_i}{(\gamma-1)}, U_f=\frac{P V_f}{(\gamma-1)}}

Therefore, change in internal energy is

\mathrm{\begin{aligned} & \Delta U=U_f-U_i=\frac{P}{(\gamma-1)}\left(V_f-V_{\mathrm{i}}\right) \\ & =\frac{1 \times 10^5 \times(25-5) \times 10^{-3}}{(1.4-1)}=5000 \mathrm{~J} \end{aligned}}

From the first law of thermodynamics, the heat energy supplied to the gas is
?Q = ?W + ?U = 2000 + 5000
= 7000 J

Posted by

Anam Khan

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