To what temperature must 20.2 g of hydrogen gas be heated at 110 kPa to occupy a volume of 4.45 x 10^-2 kL

59K

Explanation:

Using the general gas law equation:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas constant (0.0821 Latm/molK)

T = temperature (K)

According to the information given in the question;

- mass of hydrogen gas = 20.2g

mole = mass/molar mass

mole = 20.2/2.015

mole = 10.01mol

- Pressure = 110 Kpa

1 kilopascal = 0.00987 atmosphere

110kPa = 110 × 0.00987

= 1.09atm

- Volume = 4.45 x 10^-2 kL

1 kL = 1000L

4.45 x 10^-2 kL = 4.45 x 10^-2 × 10³

4.45 × 10¹

= 44.5L

Using PV = nRT

T = PV/nR

T = 1.09 × 44.5/0.0821 × 10.01

T = 48.505/0.821

T = 59.02

T = 59K

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