A)5) 1 mole of an ideal gas is kept in a cylinder with a movable frict

1.	A)5) 1 mole of an ideal gas is kept in a cylinder with a movable frictionless andmassless piston at pressure of 1.0mPa, and temperature 27°C. It is then expanded tilits volume is doubled. How much work is done if the expansion is isobaric?a) 49 kJb) 4.9 kJc) 490 kJd) 0.49 k
Answer» Correct Question : 2 mole of an ideal gas is KEPT in a cylinder with a movable frictionless and massless piston at pressure of 1.0mPa, and temperature 27°C $.$ It is then expanded til its volume is doubled. How much work is done if the expansion is isobaric ? Answer : ★ Work done in isobaric process is given by W = P ∆V W denotes work done P denotes pressure ∆V denotes change in volume ★ Ideal gas equation : PV = nRT Temperature = 27°C = 300K ➝ 10‾³ × V = 2 × 8.314 × 300 ➝ V = 4988.4 × 10³ m³ ..... (i) Now work done in expansion ; ➝ W = P (V₂ - V₁) ATQ : V₂ = 2V ➝ W = P (2V - V) ➝ W = P × V From equation (i) ➝ W = 10‾³ × (4988.4 × 10³) ➝ W = 4.9 kJ ∴ (B) is the correct answer! ★ Additional Information : The thermodynamic state of a system is characterized by its INTERNAL energy, not heat. That is internal energy is a state variable. For this REASON, the statement like a gas in given state has a certain amount of energy is meaningful.

A)5) 1 mole of an ideal gas is kept in a cylinder with a movable frictionless andmassless piston at pressure of 1.0mPa, and temperature 27°C. It is then expanded tilits volume is doubled. How much work is done if the expansion is isobaric?a) 49 kJb) 4.9 kJc) 490 kJd) 0.49 k

Answer»

Correct Question :

2 mole of an ideal gas is KEPT in a cylinder with a movable frictionless and massless piston at pressure of 1.0mPa, and temperature 27°C $.$ It is then expanded til

its volume is doubled. How much work is done if the expansion is isobaric ?

Answer :

★ Work done in isobaric process is given by

W = P ∆V

W denotes work done

P denotes pressure

∆V denotes change in volume

★ Ideal gas equation :

PV = nRT

Temperature = 27°C = 300K

➝ 10‾³ × V = 2 × 8.314 × 300

➝ V = 4988.4 × 10³ m³ ..... (i)

Now work done in expansion ;

➝ W = P (V₂ - V₁)

ATQ : V₂ = 2V

➝ W = P (2V - V)

➝ W = P × V

From equation (i)

➝ W = 10‾³ × (4988.4 × 10³)

➝ W = 4.9 kJ

∴ (B) is the correct answer!

★ Additional Information :

The thermodynamic state of a system is characterized by its INTERNAL energy, not heat. That is internal energy is a state variable.
For this REASON, the statement like a gas in given state has a certain amount of energy is meaningful.