Uma amostra de gás ideal é comprimida de 20m3 a 10m3 enquanto sua pressão aumenta de 10 Pa para 20 Pa, a temperatura T constante. O trabalho realizado sobre o gás é, em Joule,

Let's break down the problem step by step. We have an ideal gas that is compressed from 20m³ to 10m³, and its pressure increases from 10 Pa to 20 Pa, while the temperature T remains constant. We need to find the work done on the gas in Joules.

The correct answer is option C) -200 ln 1/2. But why?

When the gas is compressed, the work done on the gas is given by the formula W = -nRT ln(V_f/V_i), where n is the number of moles of the gas, R is the gas constant, T is the temperature, V_f is the final volume, and V_i is the initial volume.

In this case, we know that the initial volume is 20m³ and the final volume is 10m³. We also know that the temperature T is constant, but its value is not given. However, we don't need to know the value of T to find the work done on the gas.

Plugging in the values, we get W = -nRT ln(10/20) = -nRT ln(1/2). Since the temperature is constant, nRT is also constant, so we can simplify the expression to W = -K ln(1/2), where K is a constant.

Now, we can use the property of logarithms that says ln(1/x) = -ln(x). Applying this to our expression, we get W = K ln(2) = -200 ln(1/2), which is option C.

The other options can be eliminated by analyzing the units and the physical significance of the expressions. For example, option A has units of Joules, but the expression is not dimensionally correct. Option B has the correct units, but the expression is not physically meaningful in this context. Option D has units of Joules, but the expression is not dimensionally correct either.

Therefore, the correct answer is option C) -200 ln 1/2.