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Calculate the work done when 1 mol of an...

Calculate the work done when 1 mol of an ideal gas is compressed reversibly from 1 bar to 4 bar at a constant temperature of 300 K

A

4.01 kJ

B

`-8.02kJ `

C

18.02kJ

D

3.45kJ

Text Solution

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The correct Answer is:
To calculate the work done when 1 mole of an ideal gas is compressed reversibly from 1 bar to 4 bar at a constant temperature of 300 K, we can use the formula for work done in a reversible isothermal process: ### Step-by-Step Solution 1. **Identify the Given Values:** - Number of moles (n) = 1 mol - Initial pressure (P1) = 1 bar - Final pressure (P2) = 4 bar - Temperature (T) = 300 K - Gas constant (R) = 8.314 J/(mol·K) 2. **Use the Formula for Work Done:** The work done (W) during a reversible isothermal process can be calculated using the formula: \[ W = -nRT \ln\left(\frac{P_2}{P_1}\right) \] Here, since we are compressing the gas, we will take the logarithm of the ratio of the final pressure to the initial pressure. 3. **Substitute the Values into the Formula:** \[ W = -1 \times 8.314 \times 300 \times \ln\left(\frac{4}{1}\right) \] 4. **Calculate the Natural Logarithm:** \[ \ln(4) = 1.3863 \] Thus, we can rewrite the equation: \[ W = -1 \times 8.314 \times 300 \times 1.3863 \] 5. **Perform the Calculation:** \[ W = -1 \times 8.314 \times 300 \times 1.3863 \approx -3457.69 \text{ J} \] 6. **Convert the Work Done to Kilojoules:** Since the answer is required in kilojoules, we convert joules to kilojoules: \[ W = -3457.69 \text{ J} \times \frac{1 \text{ kJ}}{1000 \text{ J}} = -3.45769 \text{ kJ} \] Rounding to two decimal places, we get: \[ W \approx -3.46 \text{ kJ} \] 7. **Final Answer:** The work done when 1 mole of an ideal gas is compressed reversibly from 1 bar to 4 bar at a constant temperature of 300 K is approximately **-3.46 kJ**.
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The pressure-volume of varies thermodynamic process is shown in graphs: Work is the mole of transference of energy. It has been observed that reversible work done by the system is the maximum obtainable work. w_(rev) gt w_(irr) The works of isothermal and adiabatic processes are different from each other. w_("isothermal reversible") = - 2.303 nRT log_(10) ((V_(2))/(V_(1))) = 2.303 nRT log_(10)((P_(2))/(P_(1))) w_("adiabatic reversible") = C_(V) (T_(1)-T_(2)) Calculate work done when 1 mole of an ideal gas is expanded reversibly from 30L to 60L at a constant temperature of 300K

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Knowledge Check

  • What is the work done when 1 mole of a gas expands isothermally from 25 L to 250 L at a constant pressure of 1 atm and a temperature of 300 K ?

    A
    (a) `-3542 J`
    B
    (b) `-5744J`
    C
    (c) `-2657J`
    D
    (d) `-4890J`
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    Calculate the work done when 1 mole of a gas expands reversibly and isothermally from 5 atm to 1 atm at 300 K. [Value of log 5 = 0.6989].

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    The pressure-volume of varies thermodynamic process is shown in graphs: Work is the mole of transference of energy. It has been observed that reversible work done by the system is the maximum obtainable work. w_(rev) gt w_(irr) The works of isothermal and adiabatic processes are different from each other. w_("isothermal reversible") = 2.303 nRT log_(10) ((V_(2))/(V_(1))) = 2.303 nRT log_(10)((P_(2))/(P_(1))) w_("adiabatic reversible") = C_(V) (T_(1)-T_(2)) The q value and work done in isothermal reversible expansion of one mole of an ideal gas from initial pressure of 1 bar to final pressure of 0.1 bar at constant temperature 273K are: