Can you explain the answer of this question below:At 270C one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atm to 10 atm. The value of ΔE and q are (R = 2):A:0, -965.84 calB:-965.84 cal, -865.58 cal C:865.58 cal, -865.58 calD:-865.58 cal, -865.58 calThe answer is a.

Question

Can you explain the answer of this question below:At 270C one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atm to 10 atm. The value of &#916;E and q are (R = 2):A:0, -965.84 calB:-965.84 cal, -865.58 cal    C:865.58 cal, -865.58 calD:-865.58 cal, -865.58 calThe answer is a.

Answer

Isothermally (at constant temperature) and reversible work.

w = -2.303nrt log(P₁/P₂)

at constant temperature , ΔE = 0
ΔE = Q + W,

Q = −W = −965.84 cal

Answer

For the isothermal process - temperature remains constant therefore delE=0now for the q - work done on working substance because of isothermal compression. therefore workdone = heat rejected W = - Q OR Q = - W........... (1) given that P1=2 atm ; P2= 10 atm ; T = 27+ 273= 300 k we know that - PV =nRT here R and T are contant ; n=1 (according to question)therefore P proportional to 1/ Vafter using given material we got W= - 965.7 cal = Q

Answer

C

Answer

To find the value of the change in entropy, we can use the formula:

ΔS = nR ln(Vf/Vi)

where ΔS is the change in entropy, n is the number of moles of gas, R is the ideal gas constant, and Vf and Vi are the final and initial volumes of the gas, respectively.

Since the gas is compressed isothermally and reversibly, we know that the temperature remains constant throughout the process. Therefore, we can use the ideal gas law to relate the initial and final volumes and pressures:

PiVi = PfVf

where Pi and Pf are the initial and final pressures, respectively.

Rearranging this equation, we find:

Vi/Vf = Pf/Pi

Substituting this into the equation for ΔS, we have:

ΔS = nR ln(Pi/Pf)

Given that n = 1 mole, R = 0.0821 L·atm/(mol·K), Pi = 2 atm, and Pf = 10 atm, we can calculate ΔS:

ΔS = (1 mole)(0.0821 L·atm/(mol·K)) ln(2 atm/10 atm)

ΔS = (0.0821 L·atm/K) ln(0.2)

Using a calculator, we find:

ΔS ≈ -0.122 L·atm/K

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