At `27^@C` two moles of an ideal monatomic gas occupy a volume V. The gas expands adiabatically to a volume `2V`. Calculate (a) final temperature of the gas (b) change in its internal energy and (c) the work done by the gas during the process. [ `R=8.31J//mol-K`]

At `27^@C` two moles of an ideal monatomic gas occupy a volume V. The

1.	At `27^@C` two moles of an ideal monatomic gas occupy a volume V. The gas expands adiabatically to a volume `2V`. Calculate (a) final temperature of the gas (b) change in its internal energy and (c) the work done by the gas during the process. [ `R=8.31J//mol-K`]
Answer» Correct Answer - A::B::C (a) In case of adiabatic change `TV^(gamma-1)=` constant So that `T_1V_1^(gamma-1)=T_2V_2^(gamma-1)` with `gamma=(5/3)` i.e. `300xxV^(2//3)=T(2V)^(2//3)` or `T=(300)/((2)^(2//3))=189K` (b) As `DeltaU=nC_VDeltaT=n(3/2R)DeltaT So, `DeltaU=2xx(3/2)xx8.31(189-300)` `=-2767.23J` Negative sign means internal energy will decrease. (c) According to first law of thermodynamics `Q=DeltaU+DeltaW` And as for adiabatic change `DeltaQ=0` `DeltaW=-DeltaU=2767.23J`

Discussion

No Comment Found

Related InterviewSolutions

A quantity of air is kept in a container having walls which are slightly conducting. The initial temperature and volume are `27^0C` (equal to the temperature of the surrounding) and `800cm^(3)` respectively. Find the rise in the temperature of the gas is compressed to `200 cm^(3)` (a) in a short time (b) in a long time . Take gamma= `1.4`.
A gas is contained in a metallic cylinder fitted with a piston.the piston is suddenly moved in to compress the gas and is maintained at this position. As time passes the pressure of the gas in the cylinderA. increaseB. decreaseC. remains constantD. increase or decrease depending on the nature of the gas.
Five moles of an ideal monoatomic gas with an initial temperature of `127^@C` expand and in the process absorb 1200J of heat and do 2100J of work. What is the final temperature of the gas?
The ratio `(C_p)/(C_v)=gamma` for a gas. Its molecular weight is M. Its specific heat capacity at constant pressure isA. `(R)/(gamma-1)`B. `(gammaR)/(gamma-1)`C. `(gammaR)/(M(gamma-1))`D. `(gammaRM)/(gamma-1)`
A monoatomic gas of n-moles is heated temperature `T_1` to `T_2` under two different conditions (i) at constant volume and (ii) At constant pressure The change in internal energy of the gas isA. More for (i)B. More for (ii)C. Same in both casesD. Independent of number of moles
One mole of an ideal gas goes from an initial state A to final state B via two processs : It first undergoes isothermal expansion from volume `V` to `3 V` and then its volume is reduced from `3 V` to `V` at constant pressure. The correct `P-V` diagram representing the two process in (figure)A. B. C. D.
The first law of theromodynamics is a statement ofA. conservation of heatB. conservation of workC. conservation of momentumD. conservation of energy
A monatomic gas expands at constant pressure on heating. The percentage of heat supplied that increases the internal energy of the gas and that is involed in the expansion isA. `75%,25%`B. `25%,75%`C. `60%,40%`D. `40%,60%`
Two moles of an ideal monoatomic gas are expanded according to the equation pT=constant form its initial state `(p_0, V_0)` to the final state due to which its pressure becomes half of the initial pressure. The change in internal energy is A. (a) `(3p_0V_0)/(4)`B. (b) `(3p_0V_0)/(2)`C. (c) `(9p_0V_0)/(2)`D. (d) `(5p_0V_0)/(2)`
If the ratio of specific heat of a gas of constant pressure to that at constant volume is `gamma`, the change in internal energy of the mass of gas, when the volume changes from `V` to `2V` at constant pressure `p` isA. `(R)/(gamma-1)`B. `PV`C. `(PV)/((gamma-1))K`D. `(T-4)K`

At `27^@C` two moles of an ideal monatomic gas occupy a volume V. The gas expands adiabatically to a volume `2V`. Calculate (a) final temperature of the gas (b) change in its internal energy and (c) the work done by the gas during the process. [ `R=8.31J//mol-K`]

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment