0.1 mole of an ideal gas has volume `1 dm^3` in a locked box with friction less piston. The gas is in thermal equilibrium with excess of 0.5 m aqueous ethylene glycol at its freezing point. If piston is released all of a sudden at 1 atm then determine the final volume of gas in `dm^3` `(R = 0.08 atm L mol^(–1) K ^(–1) K_f = 2.0 K molal ^(–1) )`.

Calculate the temperature of 4.0 mol of a gas occupying 5 dm^3 at 3.32 bar ( R = 0.083 " bar " dm^3 K^(-1)mol^(-1) ).

Calculate the temperature of 4.0 mol of a gas occupying d dm^(3) at 3.32 bar. (R=0.083 bar dm^(3) K^(-1)mol^(-1)) .

A cylinder containing an ideal gas (0.1 mol of 1.0 dm^(3) ) is in thermal equilibrium with a large volume of 0.5 molal aqueous solution of ethylene glycol at its freezing point. If the stoppers S_(1) and S_(2) (as shown in the figure) are suddenly withdrawn, the volume of the gas in litres after equilibrium is achieved will be _____________. (Given K_(f) (water) =2.0 K kg "mol"^(-1), R=0.08 dm^(3) atm K^(-1) "mol"^(-1) )

Calculate the temperature of 5.0 mol of a gas occupying 5 dm^(3) at 3.32 bar pressure. (R = 0.083 bar dm^(3) K^(-1) mol^(-1) )

Calculate the temperature of 2.0 mol of a gas occupying 3 dm^(3) at 3.32 bar pressure. (R = 0083 bar dm^(3) K^(-1) mol^(-1) )

Calculate the temperature of 4.0 mol of a gas occupying 5 dm^(3) at 3.32 bar pressure. (R = 0083 bar dm^(3) K^(-1) mol^(-1) )

8.2 L of an ideal gas weights 9.0 g at 300 K and 1 atm pressure. The molecular mass of the gas is

′X′mol of CO_2 and 0.05 mol of Ar are enclosed in a vessel of capacity 5 L at 1atm and 27^o C .The value of ′ X ′ is (R=0.0821 atm mol^(−1) K^(−1) )

Two moles of an ideal gas at 1 atm are compressed to 2 atm at 273 K. The enthalpy change for the process is

One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1l to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300K then total entropy change of system in the above process is: [R =0.082L atm mol^(-1)K^(-1)= 8.3J mol^(-1) K^(-1) ]