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Calculate the amount in moles of a gas which occupies 10.5 dm3 at 6 atm and 30oC...


Question

Calculate the amount in moles of a gas which occupies 10.5 dm3 at 6 atm and 30oC [R = 082 atm dm3 K-1 mol-1]

Options

A) 2.536

B) 1.623

C) 4.736

D) 0.394


The correct answer is A.

Explanation:

For an ideal gas PV = nRT

Amount in moles = n

Volume v = 10.5dm3

Pressure P = 6atm

Temperature T = 30°C + 273 = 303k

R, Gas constant = 0.082 atmdm3k-1 mol

Recall from ideal gas equation

pv = nRT

n = \(\frac{RV}{RT}\)

n = \(\frac{6 \times 105}{0.082 \times 303}\)

n= 2.536mol


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Discussion (2)

  • Pressure [P]= 6 atm
    Volume [V]= \( 10.5dm^3 \)
    n= number of moles
    Gas Constant [R]= \( 0.082atm dm^{3}K^{-1}mol^{-1} \)
    Temperature [T]=30°c=30+273K=303K
    Using the Ideal Gas equation;
    PV=nRT
    n=\( \frac{PV}{RT} \)
    n= \( \frac{6× 10.5}{0.082×303} \)
    n=\( \frac{63}{24.846} \)
    n=2.5356moles
    n≈2.536 moles
    Option [A] is 💯% correct.

  • For an ideal gas PV = nRT

    Amount in moles = n

    Volume v = 10.5dm3

    Pressure P = 6atm

    Temperature T = 30°C + 273 = 303k

    R, Gas constant = 0.082 atmdm3k-1 mol

    Recall from ideal gas equation

    pv = nRT

    n = \(\frac{RV}{RT}\)

    n = \(\frac{6 \times 105}{0.082 \times 303}\)

    n= 2.536mol

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