Answer to Question #180916 in Mechanical Engineering for ajay

Question

The gas equation is PV = nRT. If P the pressure is 2.0 atm, V the volume of the gas is 6 m3, R the gas constant is 0.08206 m3 atm mole-1 K-1 and T is 300 degrees Kelvin, what are the units of n and what is its numerical value? 

Answer 1

Q180916

The gas equation is PV = nRT. If P the pressure is 2.0 atm, V the volume of the gas is 6 m³, R the gas constant is 0.08206 m³ atm mole^-1 K^-1 and T is 300 degrees Kelvin, what are the units of n and what is its numerical value? 

Solution:

please check the value of gas constant, R in the question again

I will first solve the problem using the information given in the problem.

Next I will solve the problem using correct value of 'R' .

Pressure, P = 2.0 atm.

Volume, V = 6.0 m³.

Gas constant, R = 0.08206 m³.atm/mol.K

Temperature, T = 300 K

plug all this information in the ideal gas equation, PV = nRT.

$PV = nRT$

$2.0 \ atm * \ 6.0 \ m^3 = n \ * \ 0.08206 \ m^3.atm/mol.K \ * \ 300K$

$12.0 \ m^3 \ atm = n \ * \ 24.618 m^3.atm/mol$

$n = \frac{12.0 \ m^3 \ atm}{24.618 m^3.atm/mol } = 0.487 \ mol.$

The units of n is 'mol' and the numerical value is 0.487.

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I will suggest you to check the question again.

The proper value of Gas constant ' R' is 8.206 *10^-5 m³.atm/mol.K

Now I will solve the problem, using R = 8.206 *10^-5 m³.atm/mol.K

Pressure, P = 2.0 atm.

Volume, V = 6.0 m³.

Gas constant, R = 8.206 *10^-5 m³.atm/mol.K

Temperature, T = 300 K

plug all this information in the ideal gas equation, PV = nRT.

$PV = nRT$

$2.0 \ atm * \ 6.0 \ m^3 = n \ * \ 8.206 *10^{-5} \ m^3.atm/mol.K \ * \ 300K$

$12.0 \ m^3 \ atm = n \ * \ 0.024618 m^3.atm/mol$

$n = \frac{12.0 \ m^3 \ atm}{0.024618 m^3.atm/mol } = 487 \ mol.$

If you use the correct value of Gas constant 'R' then we will get

The units of n is 'mol' and the numerical value is 487.