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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]

A) 2.536

B) 1.623

C) 4.736

D) 0.394

### 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

## 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