Question

An ideal gas at pressure Po and
temperature To undergoes a
reversible isothermal compression
and attains a pressure P1. the
characteristic gas constant is R. Net
het transferred during this process is​

Answer 1

Answer:

0

because po is postman pressure is more

Answer 2

Answer:

The net heat transferred during the process of reversible isothermal compression is, $Q = -nRT ln\frac{P_{1} }{P_{0} }$.

Explanation:

Given,

The initial pressure of an ideal gas = P₀

The temperature of an ideal gas = T₀

After the reversible isothermal compression;

The final pressure of an ideal gas becomes = P₁

The gas constant = R

The net heat transferred during the process of reversible isothermal compression =?

Now,

• From the first law of thermodynamics ;
• $\Delta U = Q- W$

Here,

• ΔU = Change in the internal energy
• Q = The net heat transfer
• W = The work done for a reversible isothermal process

As we know,

• For an isothermal process, the temperature remains constant.
• If the temperature is constant, the internal energy is also constant, i.e., ΔU = 0.

Hence,

• Q = W    -------equation (1)

Also,

• PV = constant
• P = $\frac{constant}{V}$
• $P \alpha \frac{1}{V}$

As we know,

• Work done for a reversible isothermal process is given as:
• $W = nRT ln\frac{V_{2} }{V_{1} }$

Here,

• W = Work done
• n = Number of moles of an ideal gas
• R = Gas constant
• T = Temperature
• V₂ = Final volume
• V₁ = Initial volume

Now, $P \alpha \frac{1}{V}$

Therefore, in terms of a given pressure;

• $W = nRT ln\frac{P_{0} }{P_{1} }$

Here,

• P₀ = Initial pressure
• P₁ = Final pressure

Also,

• $W = Q = -nRT ln\frac{P_{1} }{P_{0} }$

Hence, the net heat transferred during the process, $Q = -nRT ln\frac{P_{1} }{P_{0} }$.