Question

300 J of energy is given to a gas at constant volume which increases its temperature from $20^\circ$C to $50^\circ$C. If $R = 8.3$ S.I. units and $C_v = \dfrac{5R}{2}$, find mass of gas.

Hint:

In isochoric processes, supplied heat only changes internal energy since work done is zero.

Answer 1

Correct Answer: 4

Step 1: Use the relation between heat and change in internal energy

For any ideal gas, the heat supplied in a constant volume process goes entirely into changing the internal energy:

Q = ΔU

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Step 2: Write expression for change in internal energy

For an ideal gas:

ΔU = n C_v (T₂ − T₁)

Given for this gas:

C_v = 5R / 2

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Step 3: Substitute the temperature change

T₁ = 20 K,   T₂ = 50 K

ΔT = 50 − 20 = 30 K

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Step 4: Substitute numerical values

300 = n × (5R / 2) × 30

300 = n × 75R

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Step 5: Solve for number of moles

n = 300 / (75R)

n = 4 / R

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Step 6: Expression for mass of the gas

Mass of gas = n × (molecular weight)

= (4 / R) × (molecular weight)

(Molecular weight is not provided)

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Final Answer:

Number of moles of the gas:
n = 4 / R