Question

An ideal monatomic gas is confined in a cylinder by a spring loaded with the massless piston of cross-section 8 × 10^-3 m². Piston can slide on the walls of the cylinder without any friction. Initially the gas is at 300 K and occupies a volume of 2.4 × 10^-3m³ and the spring is in its relaxed position. Now, gas is slowly heated by a small electric heater and the piston moves out slowly by 0.1 m. Calculate the final temperature of the gas and the heat supplied by the heater. The force constant of the spring is 8000 N/m and atmospheric pressure 1 × 10⁵ N/m².

• A. T_F = 600 K: Q = 680 J
• B. T_F = 800 K: Q = 600 J
• C. T_F = 600 K: Q = 720 J
• D. T_F = 800 K: Q = 720 J,

Answer 1

Answer: (D)

As here initially,
P_I = P₀ = 1 × 10⁵ N/m², V_I = V₀ = 2.4 × 10^-3 m³,
T₁ = T₀ = 300 K
​And finally,
${\text{P}}_{\text{F}}={\text{P}}_0+\frac{\text{K}x}{\text{A}}=1\times {\text{10}}^5+\frac{\text{8000}\times \text{0.1}}{8\times {\text{10}}^{-3}}$
= 2 × 10⁵ N/m²
V_F = V₀ + Ax = 2.4 × 10^-3 + 0.1 × 8 × 10^-3
= 3.2 × 10^-3 m³

So from gas equation PV = nRT, i.e.,
$\frac{{\text{P}}_I{\text{V}}_I}{{\text{T}}_I}=\frac{{\text{P}}_{\text{F}}{\text{V}}_{\text{F}}}{{\text{T}}_{\text{F}}}$
$\text{we have}{\text{T}}_{\text{F}}=\frac{{\text{P}}_{\text{F}}}{{\text{P}}_{\text{I}}}\times \frac{{\text{V}}_{\text{F}}}{{\text{V}}_{\text{I}}}\times {\text{T}}_{\text{I}}$
$=\frac{2\times {\text{10}}^5}{1\times {\text{10}}^5}\times \frac{\text{3.2}\times {\text{10}}^{-3}}{\text{2.4}\times {\text{10}}^{-3}}\times \text{300}$
T_F = 800 K
As $\Delta \text{W}=\int \text{pdv}\text{,}$ here $\text{P}={\text{P}}_0+\frac{\text{Kx}}{\text{A}}$
So $\Delta \text{W}={\int }_0^{0\text{.}1}\left({\left(\text{P}\right)}_0+\frac{\text{Kx}}{\text{A}}\right)\text{Adx}={\int }_0^{0\text{.}1}\left({\left(\text{P}\right)}_0\text{A}+\text{Kx}\right)\text{dx}$
i.e., $\Delta \text{W}={\text{P}}_0\text{Ax}+\frac{1}{2}{\text{Kx}}^2$
$=\left[10^5\times 8\times 10^{-3}\times 0\text{.}1+\frac{1}{2}\times 8000\times {\left(0\text{.}1\right)}^2\right]$
i.e., $\Delta \text{W}=80+40=120\text{J}$
Note : 80 J of work is done against atmosphere and 40 J against spring.
and as $\Delta \text{U}=n{\left(\text{C}\right)}_{\text{v}}\Delta \text{T}=\frac{n\text{R}\Delta \text{T}}{\left(\gamma -1\right)}=\frac{\left({\left(\text{P}\right)}_{\text{F}}{\left(\text{V}\right)}_{\text{F}}-{\left(\text{P}\right)}_{\text{I}}{\left(\text{V}\right)}_{\text{I}}\right)}{\left(\gamma -1\right)}$
$\Delta \text{U}=\frac{\left[\left(2\times 10^5\right)\times \left(3\text{.}2\times 10^{-5}\right)-\left(1\times 10^5\right)\times \left(2\text{.}4\times 10^{-3}\right)\right]}{\left(\frac{5}{3}-1\right)}$
$\Delta \text{U}=\frac{3}{2}\times 10^2\left[6\text{.}4-2\text{.}4\right]=600\text{J}$
Hence total heat supplied,
$\Delta \text{Q}=\Delta \text{U}+\Delta \text{W}=600+120=720\text{J}$
Note : In this problem none of the variables F, P, V, T, U or Q is constant.