JEE MAIN - Physics (2021 - 26th August Evening Shift - No. 4)
A cylindrical container of volume 4.0 $$\times$$ 10$$-$$3 m3 contains one mole of hydrogen and two moles of carbon dioxide. Assume the temperature of the mixture is 400 K. The pressure of the mixture of gases is :
[Take gas constant as 8.3 J mol$$-$$1 K$$-$$1]
[Take gas constant as 8.3 J mol$$-$$1 K$$-$$1]
249 $$\times$$ 101 Pa
24.9 $$\times$$ 103 Pa
24.9 $$\times$$ 105 Pa
24.9 Pa
Explanation
V = 4 $$\times$$ 10$$-$$3 m3
n = 3 moles
T = 400 K
PV = nRT $$\Rightarrow$$ P = $${{nRT} \over V}$$
P = $${{3 \times 8.3 \times 400} \over {4 \times {{10}^{ - 3}}}}$$
= 24.9 $$\times$$ 105 Pa
n = 3 moles
T = 400 K
PV = nRT $$\Rightarrow$$ P = $${{nRT} \over V}$$
P = $${{3 \times 8.3 \times 400} \over {4 \times {{10}^{ - 3}}}}$$
= 24.9 $$\times$$ 105 Pa
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