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1 g of helium gas is confined in a two-l...

1 g of helium gas is confined in a two-litre flask unser a pressure of 2.05 atm. What is temperature ?

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The correct Answer is:
200 K
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Knowledge Check

  • 28 g "of" N_2 gas is contained in a flask at a pressure of 10 atm and at a temperature of 57^@ C . It is found that due to leakage in the flask, the pressure is reduced to half and the temperature to 27^@ C . The quantity of N_(2) gas that leaked out is.

    A
    `11//20 g`
    B
    `20//11 g`
    C
    `5//63 g`
    D
    `63//5 g`

  • The mass of 1 litre of helium under a pressure of 2 atm and at a temperature of 27^(@)C is

    A
    `0.16 g`
    B
    `0.32 g`
    C
    `0.48 g`
    D
    `0.65 g`

  • When 4 g of an ideal gas A is introduced into an evacuated flask kept at 25^(@)C , the pressure is found to be one atmosphere. If 6 g of another ideal gas B is then added to the same flask, the pressure becomes 2 atm at same temperature. The ratio of molecular weights (M_(A) : M_(B)) of the two gases would be

    A
    `1:2`
    B
    `2:1`
    C
    `2:3`
    D
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    Using van der Waals equation, calculate the constant a when 2 mol of a gas confined in a 4 L flasks exerts a pressure of 11.0 atm at a temperature of 300 K . The value of b is 0.05 L mol^(-1) .

    Using van der Waals equation, calculate the constant a when 2 mol of a gas confined in a 4 L flasks exerts a pressure of 11.0 atm at a temperature of 300 K . The value of b is 0.05 L mol^(-1) .

    Using van der Waal's equation calculate the constant 'a' when 2 mole of a gas confined in a 4 L flask exerts a pressure of 11.0 atm at a temperature of 300 K. The value of 'b' is 0.05 L mol^(-1) .

    Two moles of a real gas confined in a 5 L flask exerts a pressure 9.1 atm at a temperature of 27^(@)C . Calculate the value of 'a' given the value of b is "0.052 L mol"^(-1) .

    A sample of helium gas is at a temperature of 300 K and a pressure of 0.5 atm . What is the average kinetic energy of a molecule of a gas ?

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