Thermodynamics Test

Result

Thermodynamics Test - 47

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Weekly Quiz Competition

Question 1
4 / -1

A thermodynamic system goes from states (i) P₁ V to 2P₁, V (ii) P, V₁ to P, 2V₁. Then, work done in the two cases is

A
Zero, Zero

B
Zero, PV₁

C
PV₁, Zero

D
PV₁, P₁V₁

Solution
Question 2
4 / -1

If R = Universal gas constant, the amount of heat needed to raise the temperature of 2 moles of an ideal monoatomic gas from 273 K to 373 K when no work is done

A
100 R

B
150 R

C
300 R

D
500 R

Solution
Question 3
4 / -1

A system performs work ∆W when an amount of heat is ∆Q added to the system, the corresponding change in the internal energy is ∆U. A unique function of the initial and final states (irrespective of the mode of change) is

A
∆Q

B
∆W

C
∆U and ∆Q

D
∆U
Question 4
4 / -1

When the amount of work done is 333 cal and change in internal energy is 167 cal, then the heat supplied is

A
166 cal

B
333 cal

C
500 cal

D
400 cal

Solution
Question 5
4 / -1

First law of thermodynamics states that

A
System can do work

B
System has temperature

C
System has pressure

D
Heat is a form of energy
Question 6
4 / -1

A thermo-dynamical system is changed from state (P₁, V₁) to (P₂, V₂) by two different process. The quantity which will remain same will be

A
∆Q

B
∆W

C
∆Q + ∆W

D
∆Q – ∆W

Solution
Question 7
4 / -1

Which of the following statement is true?

A
Internal energy of a gas depends only on the state of the gas.

B
In an isothermal process change in internal energy is maximum.

C
Area under pressure, volume graph equals heat supplied in any process.

D
Work done is state dependent but not path dependent
Question 8
4 / -1

A perfect gas contained in a cylinder is kept in vacuum. If the cylinder suddenly bursts, then the temperature of the gas

A
Remains constant

B
Becomes zero

C
Increases

D
Decreases
Question 9
4 / -1

If ∆Q and ∆W represent the heat supplied to the system and the work done on the system respectively, then the first law of thermodynamics can be written as

A

B

C

D

Solution
Question 10
4 / -1

In a given process for an ideal gas, dW = 0 and DQ < 0. Then, for the gas

A
The temperature will decrease

B
The volume will increase

C
The pressure will remain constant

D
The temperature will increase

Solution