Answer
Verified
110.7k+ views
Hint: Use the first law of thermodynamics which states that, if the quantity of heat supplied to the system is capable of doing work, then the quantity of heat absorbed by the system is equal to the sum of the external work done by the system, and the increase in the internal energy of the system. Mathematically,
$dQ = dW + dU$
Complete step by step solution
According to the first law of thermodynamics,
$dQ = dW + dU$ ……(i)
Where, dQ = Amount of heat added to the system.
dW = External work done by the system.
dU = Change in internal energy of the system.
Now, we know that specific heat of a gas at constant volume \[\left( {{C_v}} \right)\] is defined as the amount of heat required to raise the temperature of 1g gas through \[1^\circ C\] keeping the volume of the gas constant.
${C_v} = {\left( {\dfrac{{dQ}}{{dt}}} \right)_v} = {\left( {\dfrac{{dU}}{{dt}}} \right)_v}$ ……(ii)
Again, \[dW = PdV\] ……(iii)
Where, P = Pressure
dV = Change in volume
As volume is constant,
$dV = 0$
$\therefore dW = 0$, (From equation (iii))
So, equation (i) becomes
$dQ = dU$
Again, using equation (ii), we get
$dU = {C_v}dT$
For n mole of gas,
$dU = n{C_v}dT$
This can only happen if the temperature change occurred at constant volume. Therefore correct option is A
Note: In thermodynamics, state function is the property whose value does not depend on the path taken by the system to reach a specific value. For example, if a system changes from state 1 to state 2 then the value of dU will depend on the value of dT at state 1 and 2 but not on the path taken to reach the desired result.
$dQ = dW + dU$
Complete step by step solution
According to the first law of thermodynamics,
$dQ = dW + dU$ ……(i)
Where, dQ = Amount of heat added to the system.
dW = External work done by the system.
dU = Change in internal energy of the system.
Now, we know that specific heat of a gas at constant volume \[\left( {{C_v}} \right)\] is defined as the amount of heat required to raise the temperature of 1g gas through \[1^\circ C\] keeping the volume of the gas constant.
${C_v} = {\left( {\dfrac{{dQ}}{{dt}}} \right)_v} = {\left( {\dfrac{{dU}}{{dt}}} \right)_v}$ ……(ii)
Again, \[dW = PdV\] ……(iii)
Where, P = Pressure
dV = Change in volume
As volume is constant,
$dV = 0$
$\therefore dW = 0$, (From equation (iii))
So, equation (i) becomes
$dQ = dU$
Again, using equation (ii), we get
$dU = {C_v}dT$
For n mole of gas,
$dU = n{C_v}dT$
This can only happen if the temperature change occurred at constant volume. Therefore correct option is A
Note: In thermodynamics, state function is the property whose value does not depend on the path taken by the system to reach a specific value. For example, if a system changes from state 1 to state 2 then the value of dU will depend on the value of dT at state 1 and 2 but not on the path taken to reach the desired result.
Recently Updated Pages
Write an article on the need and importance of sports class 10 english JEE_Main
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
If x2 hx 21 0x2 3hx + 35 0h 0 has a common root then class 10 maths JEE_Main
The radius of a sector is 12 cm and the angle is 120circ class 10 maths JEE_Main
For what value of x function fleft x right x4 4x3 + class 10 maths JEE_Main
Other Pages
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
The energy stored is a condenser is in the form of class 12 physics JEE_Main
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
In Searles apparatus when the experimental wire is class 11 physics JEE_Main