Question

An ideal monoatomic gas of two moles occupying a volume V at$27°C$.Gas expanding adiabatically having volume $2V.$calculate (a)the final temperature of the gas and (b) change in its internal energy.

• A. (a) $189$ K (b) $-2.7$kJ
• B. (a) $195$ K (b) $2.7$ kJ
• C. (a) $189$ K (b) $2.7$ kJ
• D. (a) $195$ K (b) $-2.7$ kJ

Answer 1

The correct option is A

(a) $189$ K (b) $-2.7$kJ

Explanation for the correct options:(1)(a)$189\mathrm{K}$(b)$-2.7\mathrm{kJ}$

Step 1:

• ${\mathrm{PV}}^{\mathrm{\gamma }}$=constant
• PV=nRT
• where, P is Pressure, V=Volume, R is the ideal gas constant,n is the amount of substance, and T is the temperature
• $V^{\gamma -1}$$\alpha$$\frac{1}{\mathrm{T}}$

Step 2:

• ${\left(\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}\right)}^{\mathrm{\gamma }-1}=\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right)$
• ${\mathrm{T}}_2={\mathrm{T}}_1{\left(\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}\right)}^{\mathrm{\gamma }-1}$---(i)

Step 3:

• From equation (i) ${\mathrm{T}}_2=\left(300\mathrm{K}\right){\left(\frac{\mathrm{V}}{2\mathrm{V}}\right)}^{\frac{5}{3}-1}$
• T₂$={\left(\frac{1}{2}\right)}^{\frac{5}{3}-1}$$={\left(\frac{1}{2}\right)}^{\frac{2}{3}}={\left(\frac{1}{4}\right)}^{\frac{1}{3}}=0.63$
• T₂=T₁$\times 0.63$$=300\times 0.63$$=189\mathrm{K}$

Step4:

• change in its internal energy,$∆\mathrm{U}=\frac{\mathrm{f}}{2}\mathrm{nR}∆\mathrm{T}$
• where f=degree of freedom, n is moles of gases, T is the temperature, and R is the ideal constant.
• $∆\mathrm{U}=\frac{-3}{2}\times 2\times 8.3\times 111$$=-2.76\mathrm{kJ}$

Therefore, the final temperature of the gas is $189\mathrm{K}$ and the change in its internal energy is $-2.7\mathrm{kJ}$