Question

Question 1
4.52 moles of Propane in a canister has a volume of 21.0 L and a temperature of 298.0 K what is the pressure in kPa? kPa

R=8.31 L * KPa/mole * K

Round your final answer to the nearest 10th (one decimal point)

Question 2
What is the temperature, in K, of 1.60 mol of hydrogen gas that has a volume of 42.58 liters under 950.2 mmHg of pressure? K

R=8.31 L * KPa/mole * K

101.3 KPa = 760.0 mmHg

Round your final answer to the nearest 10th (one digit after the decimal point).

Answer 1

Answer:

(1). The pressure is 533 KPa.

(2).  The temperature is 406 K.

Explanation:

Given that,

Number of moles = 4.52 moles

Volume = 21.0 L

Temperature = 298 K

(1). We need to calculate the pressure

Using ideal equation of gas

$PV=nRT$

$P=\dfrac{nRT}{V}$

Where, T = temperature

R = gas constant

V = volume

n = number of moles

Put the value into the formula

$P=\dfrac{4.52\times8.31\times298}{21.0}$

$P=533.0\ KPa$

(2). Number of moles = 1.60 moles

Volume = 21.0 L

Temperature = 298 K

We need to calculate the temperature

Using ideal equation of gas

$PV=nRT$

$T=\dfrac{PV}{nR}$

Put the value into the formula

$T=\dfrac{126.68293\times42.58}{1.60\times8.31}$

$T=406\ K$

Hence, (1). The pressure is 533 KPa.

(2).  The temperature is 406 K.