Question

A gas undergoes two processes. In the first, the volume remains constant at 0.170 m3 and the pressure increases from 1.50×105 Pa to 6.00×105 Pa . The second process is a compression to a volume of 0.130 m3 at a constant pressure of 6.00×105 Pa. Find the total work done by the gas during both processes.

Answer 1

Answer:

$W_{T} = - 24 kJ$

Explanation:

The work (W) done by the gas can be calculated using the following equation:

$W = p*\Delta V = p*(V_{f} - V_{i})$

Where:

p: is the pressure

[tex}V_{f}[/tex]: is the final volume

[tex}V_{i}[/tex]: is the initial volume

In the first process, the work done by the gas is:

$W_{1} = p*\Delta V = p*0 = 0$

Since the volume remains constant, the total work done by the gas is equal to zero.

In the second process, the work done by the gas is:

$W_{2} = p*(V_{f} - V_{i}) = 6.00 \cdot 10^{5} Pa*(0.130 m^{3} - 0.170 m^{3}) = -24 kJ$

Now, the total work done by the gas during both processes is:

$W_{T} = W_{1} + W_{2} = 0 + (-24 kJ) = - 24 kJ$

Therefore, the total work done by the gas during both processes is - 24 kJ.

I hope it helps you!

Answer 2

The total work done by the gas during both processes is 0 Joules and  -2.4 × 10⁴ Joules respectively.

The work done on a system at constant pressure (P) can be expressed by using the relation:

W = ρdV

here:

• dV = change in volume, and since the volume is said to remain constant in the first process, then V = (0)

∴

W = ρ(0)

W = 0 Joules

In the second process;

• V₂ = 0.130 m³
• V₁ = 0.170 m³

Using the same relation;

W = ρdV

where

• dV = (0.130 - 0.170) m³
• dV = - 0.04 m³

∴

W = 6.00 × 10⁵ (-0.04) Joules

W = -24000 Joules

W = -2.4 × 10⁴ Joule

Learn more about work done here:

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