# The volume of 2.8g of carbon monoxide at and 0.821atm pressure is: (R= 0.0821 lit.atm.mol-1K1):A.1.5 litB.0.3litC.3litD.30lit

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Hint:Ideal gas law is a combined set of gas laws that is a thermodynamic equation that allows us to relate the temperature, volume, and number of moles present in a sample of gas.
Formula used: The formula for calculating V is:
${\text{V = }}\dfrac{{{\text{WRT}}}}{{{\text{PM}}}}$

As we know that, the formula is ${\text{V = }}\dfrac{{{\text{WRT}}}}{{{\text{PM}}}}$
Where,
W=weight
R= constant (given in the question)
P= pressure
T=temperature
M= molar mass of carbon monoxide (i.e. 28)
Hence, by applying the formula and putting the values
We get,
$V = \dfrac{{2.8 \times 0.0821 \times 300}}{{0.821 \times 28}} \\ = \dfrac{{68.9}}{{22.9}} \\ = 3lit \\$
Thus, we get V= 3 lit

Hence, option C is correct.

Note:
Another way is,
From gas equation, PV=NRT,
$N = \dfrac{m}{M} = \dfrac{{2.8}}{{28}} = 0.1 \\ V = \dfrac{{0.821 \times 300}}{{0.1 \times 0.821}} = 3L \\$