Question

 An ideal gas of mass m is contained in a rigid tank of volume V at a pressure p. During a reversible process its pressure reduces to \(p_2\). Following statements are made regarding the process.

(P) Heat is transferred from the gas.

(Q) Work done by the gas is zero.

(R) Entropy of the gas remains constant.

(S) Entropy of the gas decreases.

Among the above statements, the correct ones are

1.

P and R only.2. P, Q and R only.<<br>3. Q and R only.4. P, Q and S only.<

Answer 1

Correct Answer - Option 4 :

P, Q a
Concept:

Work done of the gas is given as (closed system):

\({{\rm{W}}_{1 - 2}} = {\rm{\;}}\mathop \smallint \limits_1^2 {\rm{P}} × {\rm{dV}}\)

Entropy: It is a property that is the measure of the degree of randomness or disorder of the system. It is the measure of the system's thermal energy with respect to temperature.

\(dS\ =\ \frac{dQ}{T}\)

Gay-Lussac's law: It is defined as the pressure of the given mass of a gas that varies with respect to temperature at constant volume.

P ∝ T at V = Constant

Explanation:

• Based on the above law to decrease the pressure the temperature has to be decreased and the volume is constant.
• To decrease the temperature heat should be transferred from the gas.
• Heat is decreased because it is being transferred from the gas and hence the entropy of the gas decreases because from the above equation the change in heat energy is directly proportional to the change in entropy of the gas.
• Since volume is constant the change in volume is zero and hence the workdone by the gas is zero.

Work transfer for various closed system processes are:

1) W = 0 in constant volume process

2) W = P × (V₁ - V₂) in constant pressure process

3) \(W=P_1\ \times\ V_1\ \times ln\left ( \frac{V_2}{V_1} \right )\)  in the isothermal process

4) \(W=\frac{P_1V_1\ -\ P_2V_2}{\gamma\ -\ 1}\)  in an adiabatic process

5) \(W=\frac{P_1V_1\ -\ P_2V_2}{n\ -\ 1}\) in the polytropic process