Test Index

Heat and Thermodynamics

Section: Physics

Question : 58 of 69

Marks: +1, -0

T_1

Solution:

For an adiabatic process, we have

T_1 V_1^{\gamma-1}

T_2 V_2^{\gamma-1}

For monatomic gas γ = 5/3, and it is given that initial volume

V_1

= 5.6 L and temperature

T_1

. Therefore,

T_2

T_1 \left( \frac{V_1}{V_2} \right)^{\gamma-1}

T_1 \left( \frac{5.6}{0.7} \right)^{\frac{5}{3}-1}

8^{\frac{2}{3}} \times T_1

For an adiabatic process, the work done is

W =

\frac{nR}{\gamma-1} (T_2-T_1)

However, at STP, 1 mol of substance occupies 22.4 L.

Therefore,

n =

\frac{5.6}{22.4}

\frac{1}{4} \mathrm{mol}

W =

\frac{R(4T_1-T_1)}{4\left[ \frac{5}{3}-1 \right]}

\frac{9}{8} RT_1

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