मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Solve the following. Calculate the pressure in atm of 1.0 mole of helium in a 2.0 dm3 container at 20.0°C. - Chemistry

Advertisements
Advertisements

प्रश्न

Solve the following.

Calculate the pressure in atm of 1.0 mole of helium in a 2.0 dm3 container at 20.0°C.

संख्यात्मक
Advertisements

उत्तर

Given:
n = number of moles = 1.0 mol,
V = volume = 2.0 dm3
T = Temperature = 20.0°C = 20.0 + 273.15 K = 293.15 K
R = 0.0821 L atm K–1 mol–1

To find: Pressure (P)

Formula: PV = nRT

Calculation:

According to ideal gas equation,

PV = nRT

∴ P = `"nRT"/"V"`

∴ P = `(1.0xx0.0821xx293.15)/2.0`

∴ P = 12.03 atm

The pressure of the given helium gas is 12.03 atm.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 10: States of Matter - Exercises [पृष्ठ १५९]

APPEARS IN

बालभारती Chemistry [English] Standard 11 Maharashtra State Board
पाठ 10 States of Matter
Exercises | Q 5. (J) | पृष्ठ १५९

संबंधित प्रश्‍न

Answer in one sentence.

When a gas is heated the particles move more quickly. What is the change in the volume of a heated gas if the pressure is kept constant?


Solve the following.

Nitrogen gas is filled in a container of volume 2.32 L at 32°C and 4.7 atm pressure. Calculate the number of moles of the gas.


Solve the following.

Calculate the number of molecules of methane in 0.50 m3 of the gas at a pressure of 2.0 × 102 kPa and a temperature of exactly 300 K.


When an ideal gas undergoes unrestrained expansion, no cooling occurs because the molecules


A bottle of ammonia and a bottle of HCl connected through a long tube are opened simultaneously at both ends. The white ammonium chloride ring first formed will be


If the temperature and volume of an ideal gas is increased to twice its values, the initial pressure P becomes


At identical temperature and pressure, the rate of diffusion of hydrogen gas is `3sqrt3` times that of a hydrocarbon having molecular formula CnH2n–2. What is the value of n?


The variation of volume V, with temperature T, keeping the pressure constant is called the coefficient of thermal expansion ie α = `1/"V"((∂"V")/(∂"T"))_"P"`. For an ideal gas α is equal to


Suggest why there is no hydrogen (H2) in our atmosphere. Why does the moon have no atmosphere?


A combustible gas is stored in a metal tank at a pressure of 2.98 atm at 25°C. The tank can withstand a maximum pressure of 12 atm after which it will explode. The building in which the tank has been stored catches fire. Now predict whether the tank will blow up first or start melting? (Melting point of the metal = 1100 K).


For an ideal gas, at constant temperature and pressure, the volume is ____________.


At a constant pressure, an ideal gas has a volume of 200 cm3 at 25°C. If the gas is cooled to −3°C, what will be the final volume of a gas?


What mass of an oxygen gas will occupy 8.21 L of volume at 1 atm pressure and 400 K temperature?


At a constant pressure, the density of a certain amount of an ideal gas is ____________.


If two moles of an ideal gas at 546 K occupy a volume of 44.8 L. What is the pressure of ideal gas at 546 K? (R = 0.0821 L atm mol-1 K-1)


An evacuated glass vessel weighs 40 g when empty, 135 g when filled with a liquid of density 0.95 g mL−1 and 40.5 g when filled with an ideal gas at 0.82 atm at 250 K. The molar mass of the gas in g mol−1 is ______.

(Given: R = 0.082 L atm K−1 mol−1)


In Duma's method, 0.52 g of an organic compound on combustion gave 68.6 mL N2 at 27°C and 756 mm pressure. What is the percentage of nitrogen in the compound?


Gas equation, pV = nRT is obeyed by a gas in ______.


The ideal gas equation for n moles of gas is:


The value of the universal gas constant R is:


A pressure cooker works on the principle of:


A hot air balloon rises because:


According to Boyle's Law, at constant temperature, the pressure of a gas is:


An oxygen cylinder of volume 30 litres has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressure at temperature 27°C. The mass of the oxygen withdrawn from the cylinder is nearly equal to:

[Given: R = `100/12`Jmol−1 K−1,and molecular mass of 12 O2 = 32, 1 atm pressure = 1.01 × 105 N/m]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×