Question

A gas is enclosed in a cylindrical vessel fitted with a frictionless piston. The gas is slowly heated for some time. During the process, 10 J of heat is supplied and the piston is found to move out 10 cm. Find the increase in the internal energy of the gas. The area of cross section of the cylinder = 4 cm2 and the atmospheric pressure = 100 kPa.

Answer 1

The increase in the internal energy of the gas is ΔU=6 J

Explanation:

Slowly meas the pressure is constant .

Given data:

Heat supplied = 10 J

Area of cross section of the cylinder = 4 cm^2 = 4 x 10^-4 m

P = P0 =  100 kPa (atmospheric pressure)

ΔW= PΔV= P0Ax

= 100×103×4×10−4×0.1

= 4J

From first law of thermodynamics

ΔQ=ΔU+ΔW

10=ΔU+4

ΔU=6 J

Thus the increase in the internal energy of the gas is ΔU=6 J

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Answer 2

Positive sign suggests an increase in the system's internal energy.

Explanation:

Step 1:

∆Q = 10 J

∆Q   is the Heat which is supplied to the system,  

∆V   is the Change in volume of the system,

∆V = cross section area  × the piston  Displacement  

$\begin{array}{l}{=A \times 10 \mathrm{cm}} \\{=(4 \times 10) \mathrm{cm}^{3}=40 \times 10^{-6} \mathrm{m}^{3}}\end{array}$

Step 2:

P = 100 kPa

$\begin{aligned}\Delta W &=P \Delta V=100 \times 10^{3} \times 40 \times 10^{-6} \mathrm{m}^{3} \\&=4 \mathrm{J}\end{aligned}$

∆U = ?

Step 3:

Using the thermodynamics first law, we get

 $\begin{aligned}&10=\Delta U+\Delta W\\&10=\Delta U+4\end{aligned}$

  ∆U = 6 J

Here, positive sign suggests an increase in the system's internal energy.