Home
Class 11
PHYSICS
One mole of a monoatomic ideal gas is ta...

One mole of a monoatomic ideal gas is taken through the cycle ABCDA as shown in the figure.
`T_A=1000K` and `2p_A=3p_B=6p_C`

`[Ass um e(2/3)^0.4=0.85` and `R=(25)/(3)JK^-1mol^-1]`
The temperature at B is

A

350K

B

1175K

C

850K

D

577K

Text Solution

Verified by Experts

The correct Answer is:
C
In adiabatic process,
`p^(1-gamma)T^gamma=`constant or `Tpropp^((gamma-1)/(gamma))`
`:.` `(T_B)/(T_A)=((p_B)/(p_A))^((gamma-1)/(gamma)`
`:.` `T_B=(1000)(2/3)^((5//3-1)/(5//3))`
`=(1000)(2/3)^0.4=850K`
Promotional Banner

Topper's Solved these Questions

  • LAWS OF THERMODYNAMICS

    DC PANDEY|Exercise Level 2 Subjective|18 Videos

  • LAWS OF THERMODYNAMICS

    DC PANDEY|Exercise Level 2 Passage I|2 Videos

  • LAWS OF MOTION

    DC PANDEY|Exercise Medical entrances gallery|39 Videos

  • MAGNETIC EFFECT OF CURRENT AND MAGNETISM

    DC PANDEY|Exercise Integer type Questions|10 Videos

Similar Questions

Explore conceptually related problems

One mole of a monoatomic ideal gas is taken through the cycle ABCDA as shown in the figure. T_A=1000K and 2p_A=3p_B=6p_C [Ass um e(2/3)^0.4=0.85 and R=(25)/(3)JK^-1mol^-1] Work done by the gas in the process ArarrB is

One mole of a monoatomic ideal gas is taken through the cycle ABCDA as shown in the figure. T_A=1000K and 2p_A=3p_B=6p_C [Ass um e(2/3)^0.4=0.85 and R=(25)/(3)JK^-1mol^-1] Heat lost by the gas in the process BrarrC is

One mole of a monatomic ideal gas is taken through the cycle shown in the figure: A to B adiabatic expansion B to C: cooling at constant volume C to D adiabatic compression D to C: heating at constant volume. The pressure and temperature at P_A, P_B etc. T_A , T_B etc. are denoted etc. respectively. Given that T_A = 1000 K, P_B =(2 //3)P_A and P_C = (1 //3)P_A Calculate the following quantities. The work done by the gas in process A to B (in J)

One mole of a monatomic ideal gas is taken through the cycle shown in the figure: A to B adiabatic expansion B to C: cooling at constant volume C to D adiabatic compression D to C: heating at constant volume. The pressure and temperature at P_A, P_B etc. T_A , T_B etc. are denoted etc. respectively. Given that T_A = 1000 K, P_B =(2 //3)P_A and P_C = (t //3)P_A Calculate the following quantities. The heat lost by the gas in process B to C (in J)

One mole of a monatomic ideal gas is taken through the cycle shown in the figure: A to B adiabatic expansion B to C: cooling at constant volume C to D adiabatic compression D to A: heating at constant volume. The pressure and temperature at P_A, P_B etc. T_A , T_B etc. are denoted etc. respectively. Given that T_A = 1000 K, P_B =(2 //3)P_A and P_C = (1 //3)P_A Calculate the following quantities. The temperature T_D (in K) [Given : (2/5)^(2//5) = 0.85 ]

Half mole of an ideal gas (gamma = 5/3) is taken through the cycle abcda as shown in figure . Take (R = 25/3 JK^(-1) mol^(-1)) . (a) Find the temperature of the gas in the staes a, b, c and d. (b) Find the amount of heat supplied in the processes ab and bc. (c ) Find the amount of heat liberated in the processes cd and da .

One mole of a monoatomic ideal gas is taken through the cycle shown in figure. ArarrB Adiabatic expansion BrarrC Cooling at constant volume CrarrD Adiabatic compression. DrarrA Heating at constant volume The pressure and temperature at A,B etc., are denoted by p_A, T_A, p_B, T_B etc. respectively. Given, T_A=1000K , p_B=(2/3)p_A and p_C=(1/3)p_A . Calculate (a) the work done by the gas in the process ArarrB (b) the heat lost by the gas in the process BrarrC Given, (2/3)^0.4=0.85 and R=8.31J//mol-K