1
JEE Advanced 2021 Paper 2 Online
+3
-1
A thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle, as shown in the figure below. On each side of the partition, there is one mole of an ideal gas, with specific heat at constant volume, CV = 2R. Here, R is the gas constant. Initially, each side has a volume V0 and temperature T0. The left side has an electric heater, which is turned on at very low power to transfer heat Q to the gas on the left side. As a result the partition moves slowly towards the right reducing the right side volume to V0/2. Consequently, the gas temperatures on the left and the right sides become TL and TR, respectively. Ignore the changes in the temperatures of the cylinder, heater and the partition.

The value of $${{{T_R}} \over {{T_0}}}$$ is
A
$$\sqrt 2$$
B
$$\sqrt 3$$
C
2
D
3
2
JEE Advanced 2021 Paper 2 Online
+3
-1
A thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle, as shown in the figure below. On each side of the partition, there is one mole of an ideal gas, with specific heat at constant volume, CV = 2R. Here, R is the gas constant. Initially, each side has a volume V0 and temperature T0. The left side has an electric heater, which is turned on at very low power to transfer heat Q to the gas on the left side. As a result the partition moves slowly towards the right reducing the right side volume to V0/2. Consequently, the gas temperatures on the left and the right sides become TL and TR, respectively. Ignore the changes in the temperatures of the cylinder, heater and the partition.

The value of $${Q \over {R{T_0}}}$$ is
A
$$4(2\sqrt 2 + 1)$$
B
$$4(2\sqrt 2 - 1)$$
C
$$(5\sqrt 2 + 1)$$
D
$$(5\sqrt 2 - 1)$$
3
JEE Advanced 2021 Paper 1 Online
+3
-1
An ideal gas undergoes a four step cycle as shown in the P-V diagram below. During this cycle, heat is absorbed by the gas in

A
steps 1 and 2
B
steps 1 and 3
C
steps 1 and 4
D
steps 2 and 4
4
JEE Advanced 2019 Paper 2 Offline
+3
-1
In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by T$$\Delta$$X where T is temperature of the system and $$\Delta$$X is the infinitesimal change in a thermodynamic quantity X of the system. For a mole of monatomic ideal gas, $$X = {3 \over 2}R\,\ln \left( {{T \over {{T_A}}}} \right) + R\,\ln \left( {{V \over {{V_A}}}} \right)$$

Here, R is gas constant, V is volume of gas, TA and VA are constants.

The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities.

If the process on one mole of monatomic ideal gas is as shown in the TV-diagram with $${P_0}{V_0} = {1 \over 3}R{T_0}$$, the correct match is,

A
I $$\to$$ P, II $$\to$$ R, III $$\to$$ T, IV $$\to$$ S
B
I $$\to$$ P, II $$\to$$ T, III $$\to$$ Q, IV $$\to$$ T
C
I $$\to$$ S, II $$\to$$ T, III $$\to$$ Q, IV $$\to$$ U
D
I $$\to$$ P, II $$\to$$ R, III $$\to$$ T, IV $$\to$$ P
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