Question

The figure shows a cylindrical container containing oxygen (γ = 1.4) and closed by a 50-kg frictionless piston. The area of cross-section is 100 cm2, atmospheric pressure is 100 kPa and g is 10 m s−2. The cylinder is slowly heated for some time. Find the amount of heat supplied to the gas if the piston moves out through a distance of 20 cm.
Figure

Answer 1

The amount of heat supplied to the gas is

1050 J

Explanation:

Given data:

Mass of cylinder "m" = 50-kg

Cross-sectional area "A" = 100 cm^2

Pressure "P" = 100 kPa

g = 10 m s−2

Distance moved by the piston "x"  = 20 cm

Work done by the gas

dW = Pdv

Solution:

W = (mg/A + Po)  x Adx

   = ( 50 x 100/ 100 x 10^-4 + 10^5) x 100 x 10^-4 x 20 x 10^-4

   = (5 x 10^4 + 10^4) x 20 x 10^-4

   = 1.5 x 10^5  x 20 x 10^-4  

   = 300 J  =  PΔV

Thus

nRdT  = PΔV

dT = 300/nR

dQ = nCpdT = nCp x (300/nR )

dQ = nγR300/( γ-1)nR

dQ = ( 300 x 1.4/0.4) = 1050 J

The amount of heat supplied to the gas is 1050 J.

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Answer 2

The amount of heat supplied to the gas if the piston moves out through a distance of 20 cm is 1050 J.

Explanation:

Step 1:

piston  Mass (m) = 50 kg

The gas is adiabatic constant, γ = 1.4

Piston cross section area (A) = 100 $cm^2$

pressure in the atmosphere ($P_o$) = 100 kPa

g = 10 $m/s^2$  

Step 2:

The distance the piston moves, x = 20 cm

gas Work done,

dW=Pdv  

The pressure (p) is due to two factors: first is the initial pressure, and second is the piston weight.

Therefore,  

$W=\left(\frac{m g}{A}+P_{0}\right) \times A d x$

$\begin{array}{l}{W=\left(\frac{50 \times 10}{100 \times 10^{-4}}+10^{5}\right) \times 100 \times 10^{-4} \times 20 \times 10^{-4}} \\{W=\left(5 \times 10^{4}+105\right) \times 20 \times 10^{-4}} \\{W=1.5 \times 105 \times 20 \times 10^{-4}} \\{W=300 J}\end{array}$

Step 3:

Hence, nRdT = PΔV = 300

dT = 300 nR

$\text { So, } d Q=n C_{p} d T=n C_{p} \times\left(\frac{300}{n R}\right)$

$U \sin g C_{p}-C_{v}=R \text { and } \frac{C_{p}}{C_{v}}=\gamma$

$d Q=\frac{n \gamma R 300}{(\gamma-1) n R}$

$d Q=\left(\frac{300 \times 1.4}{0.4}\right)=1050 J$