What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a $$9dm^3$$ flask at 27°C ?

Asked by Pragya Singh | 1 year ago |  64

##### Solution :-

It is known that,

p =$$\dfrac{m}{M}\dfrac{RT}{V}$$

For methane (CH4),

$$p_{CH_{ 4 }}$$

$$\dfrac{ 3.2 }{ 16 }​ × \dfrac{8.314 \; \times \;300 }{9 \; \times \; 10^{-3 }}$$

$$[Since \;9dm^3 = 9 \; \times \; 10^{-3}m^3 \;27°C$$

$$= 300 K]$$

= 5.543 × $$10^{ 4 }$$ Pa

For carbon dioxide (CO2),

$$p_{CO_{ 2 }}$$

=$$\dfrac{ 4.4 }{ 44 }× \dfrac{8.314 \; \times \;300 }{9 \; \times \; 10^{-3}}$$

$$2.771 × 10^{4} Pa$$

Total pressure exerted by the mixture can be calculated as:

$$p= p_{CH_{ 4 }} + p_{CO_{ 2 }}$$

$$(5.543 × 10^{ 4 }+ 2.771 × 10^{4}) Pa$$

$$8.314 × 10^{ 4 }$$ Pa

Answered by Pragya Singh | 1 year ago

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